It can easily be verified that these sixteen triples are as required. Now let $(a, b, c)$ be any triple with the desired property. If we would have $a = 1$, then both $b - c$ and $c - b$ were powers of $2$, which is impossible since their sum is zero; because of symmetry, this argument shows $a, b, c \geqslant 2$.

Case 1. Among $a, b$, and $c$ there are at least two equal numbers.

Without loss of generality we may suppose that $a = b$. Then $a^{2} - c$ and $a(c - 1)$ are powers of $2$. The latter tells us that actually $a$ and $c - 1$ are powers of $2$. So there are nonnegative integers $\alpha$ and $\gamma$ with $a = 2^{\alpha}$ and $c = 2^{\gamma} + 1$. Since $a^{2} - c = 2^{2\alpha} - 2^{\gamma} - 1$ is a power of $2$ and thus incongruent to $-1$ modulo $4$, we must have $\gamma \leqslant 1$. Moreover, each of the terms $2^{2\alpha} - 2$ and $2^{2\alpha} - 3$ can only be a power of $2$ if $\alpha = 1$. It follows that the triple $(a, b, c)$ is either $(2, 2, 2)$ or $(2, 2, 3)$.

Case 2. The numbers $a, b$, and $c$ are distinct.

Due to symmetry we may suppose that
$$2 \leqslant a < b < c$$
We are to prove that the triple $(a, b, c)$ is either $(2, 6, 11)$ or $(3, 5, 7)$. By our hypothesis, there exist three nonnegative integers $\alpha, \beta$, and $\gamma$ such that
$$\begin{align*} bc - a & = 2^{\alpha}, \\ ac - b & = 2^{\beta}, \\ \text{and}\quad ab - c & = 2^{\gamma}. \end{align*}$$
Evidently we have
$$\alpha > \beta > \gamma.$$
Depending on how large $a$ is, we divide the argument into two further cases.

Case 2.1. $a = 2$.

We first prove that $\gamma = 0$. Assume for the sake of contradiction that $\gamma > 0$. Then $c$ is even by (4) and, similarly, $b$ is even by (5) and (3). So the left-hand side of (2) is congruent to $2$ modulo $4$, which is only possible if $bc = 4$. As this contradicts (1), we have thereby shown that $\gamma = 0$, i.e., that $c = 2b - 1$.

Now (3) yields $3b - 2 = 2^{\beta}$. Due to $b > 2$ this is only possible if $\beta \geqslant 4$. If $\beta = 4$, then we get $b = 6$ and $c = 2 \cdot 6 - 1 = 11$, which is a solution. It remains to deal with the case $\beta \geqslant 5$. Now (2) implies
$$9 \cdot 2^{\alpha} = 9b(2b - 1) - 18 = (3b - 2)(6b + 1) - 16 = 2^{\beta}(2^{\beta + 1} + 5) - 16$$
and by $\beta \geqslant 5$ the right-hand side is not divisible by $32$. Thus $\alpha \leqslant 4$ and we get a contradiction to (5).

Case 2.2. $a \geqslant 3$.

Pick an integer $\vartheta \in \{-1, +1\}$ such that $c - \vartheta$ is not divisible by $4$. Now
$$2^{\alpha} + \vartheta \cdot 2^{\beta} = (bc - a\vartheta^{2}) + \vartheta(ca - b) = (b + a\vartheta)(c - \vartheta)$$
is divisible by $2^{\beta}$ and, consequently, $b + a\vartheta$ is divisible by $2^{\beta - 1}$. On the other hand, $2^{\beta} = ac - b > (a - 1)c \geqslant 2c$ implies in view of (1) that $a$ and $b$ are smaller than $2^{\beta - 1}$. All this is only possible if $\vartheta = 1$ and $a + b = 2^{\beta - 1}$. Now (3) yields
$$ac - b = 2(a + b)$$
whence $4b > a + 3b = a(c - 1) \geqslant ab$, which in turn yields $a = 3$.

So (6) simplifies to $c = b + 2$ and (2) tells us that $b(b + 2) - 3 = (b - 1)(b + 3)$ is a power of $2$. Consequently, the factors $b - 1$ and $b + 3$ are powers of $2$ themselves. Since their difference is $4$, this is only possible if $b = 5$ and thus $c = 7$. Thereby the solution is complete.

As in the beginning of the first solution, we observe that $a, b, c \geqslant 2$. Depending on the parities of $a, b$, and $c$ we distinguish three cases.

Case 1. The numbers $a, b$, and $c$ are even.

Let $2^{A}, 2^{B}$, and $2^{C}$ be the largest powers of $2$ dividing $a, b$, and $c$ respectively. We may assume without loss of generality that $1 \leqslant A \leqslant B \leqslant C$. Now $2^{B}$ is the highest power of $2$ dividing $ac - b$, whence $ac - b = 2^{B} \leqslant b$. Similarly, we deduce $bc - a = 2^{A} \leqslant a$. Adding both estimates we get $(a + b)c \leqslant 2(a + b)$, whence $c \leqslant 2$. So $c = 2$ and thus $A = B = C = 1$; moreover, we must have had equality throughout, i.e., $a = 2^{A} = 2$ and $b = 2^{B} = 2$. We have thereby found the solution $(a, b, c) = (2, 2, 2)$.

Case 2. The numbers $a, b$, and $c$ are odd.

If any two of these numbers are equal, say $a = b$, then $ac - b = a(c - 1)$ has a nontrivial odd divisor and cannot be a power of $2$. Hence $a, b$, and $c$ are distinct. So we may assume without loss of generality that $a < b < c$.

Let $\alpha$ and $\beta$ denote the nonnegative integers for which $bc - a = 2^{\alpha}$ and $ac - b = 2^{\beta}$ hold. Clearly, we have $\alpha > \beta$, and thus $2^{\beta}$ divides
$$a \cdot 2^{\alpha} - b \cdot 2^{\beta} = a(bc - a) - b(ac - b) = b^{2} - a^{2} = (b + a)(b - a).$$
Since $a$ is odd, it is not possible that both factors $b + a$ and $b - a$ are divisible by $4$. Consequently, one of them has to be a multiple of $2^{\beta - 1}$. Hence one of the numbers $2(b + a)$ and $2(b - a)$ is divisible by $2^{\beta}$ and in either case we have
$$ac - b = 2^{\beta} \leqslant 2(a + b).$$
This in turn yields $(a - 1)b < ac - b < 4b$ and thus $a = 3$ (recall that $a$ is odd and larger than $1$). Substituting this back into (7) we learn $c \leqslant b + 2$. But due to the parity $b < c$ entails that $b + 2 \leqslant c$ holds as well. So we get $c = b + 2$ and from $bc - a = (b - 1)(b + 3)$ being a power of $2$ it follows that $b = 5$ and $c = 7$.

Case 3. Among $a, b$, and $c$ both parities occur.

Without loss of generality, we suppose that $c$ is odd and that $a \leqslant b$. We are to show that $(a, b, c)$ is either $(2, 2, 3)$ or $(2, 6, 11)$. As at least one of $a$ and $b$ is even, the expression $ab - c$ is odd; since it is also a power of $2$, we obtain
$$ab - c = 1.$$
If $a = b$, then $c = a^{2} - 1$, and from $ac - b = a(a^{2} - 2)$ being a power of $2$ it follows that both $a$ and $a^{2} - 2$ are powers of $2$, whence $a = 2$. This gives rise to the solution $(2, 2, 3)$.

We may suppose $a < b$ from now on. As usual, we let $\alpha > \beta$ denote the integers satisfying
$$2^{\alpha} = bc - a \quad \text{and} \quad 2^{\beta} = ac - b$$
If $\beta = 0$ it would follow that $ac - b = ab - c = 1$ and hence that $b = c = 1$, which is absurd. So $\beta$ and $\alpha$ are positive and consequently $a$ and $b$ are even. Substituting $c = ab - 1$ into (9) we obtain
$$\begin{align*} 2^{\alpha} & = ab^{2} - (a + b) \\ 2^{\beta} & = a^{2}b - (a + b) \end{align*}$$
The addition of both equation yields $2^{\alpha} + 2^{\beta} = (ab - 2)(a + b)$. Now $ab - 2$ is even but not divisible by $4$, so the highest power of $2$ dividing $a + b$ is $2^{\beta - 1}$. For this reason, the equations (10) and (11) show that the highest powers of $2$ dividing either of the numbers $ab^{2}$ and $a^{2}b$ is likewise $2^{\beta - 1}$. Thus there is an integer $\tau \geqslant 1$ together with odd integers $A, B$, and $C$ such that $a = 2^{\tau}A, b = 2^{\tau}B, a + b = 2^{3\tau}C$, and $\beta = 1 + 3\tau$.

Notice that $A + B = 2^{2\tau}C \geqslant 4C$. Moreover, (11) entails $A^{2}B - C = 2$. Thus $8 = 4A^{2}B - 4C \geqslant 4A^{2}B - A - B \geqslant A^{2}(3B - 1)$. Since $A$ and $B$ are odd with $A < B$, this is only possible if $A = 1$ and $B = 3$. Finally, one may conclude $C = 1, \tau = 1, a = 2, b = 6$, and $c = 11$. We have thereby found the triple $(2, 6, 11)$. This completes the discussion of the third case, and hence the solution.