Permutations of $(1, 2, 3)$ or $(x, y, y)$ where $x, y \in N$, $N$ is the set of all positive numbers.

Lemma 1. $2 < (1 + \frac{1}{n})^n < 3$ for all $n \in N_{\ge 2}$

Proof. Use binomial theorem, we have
$$2 = 1 + n \cdot \frac{1}{n} < 1 + n \cdot \frac{1}{n} + \sum_{i=2}^{n} \frac{\binom{n}{i}}{n^i} = (1 + \frac{1}{n})^n.$$
On the other hand,
$$(1 + \frac{1}{n})^n = 1 + n \cdot \frac{1}{n} + \sum_{i=2}^{n} \frac{\binom{n}{i}}{n^i} < 1 + 1 + \sum_{i=2}^{n} \frac{1}{2^{k-1}} < 3.$$

Lemma 2. $x^y > y^x$ for all $3 \le x < y \in N$.

Proof. Fix $x$, we use induction to prove this lemma. When $y = x + 1$,
$$x^{(x+1)} = x \cdot x^x \ge 3x^x > (1 + \frac{1}{x})^x \cdot x^x = (x+1)^x.$$
Suppose it's true for $y = x + 1, \dots, x + k - 1$. Then,
$$x^{x+k} = x \cdot x^{x+k-1} \ge (1 + \frac{1}{x+k-1})^x \cdot (x+k-1)^x = (x+k)^x.$$
Therefore, it's true for all $y \in N_{\ge x+1}$, as desired.

Lemma 3. If $a > b > c \ge 2$, then $a^b + b^c + c^a < a^c + b^a + c^b$

Proof. Fix $b$ and $c$, we consider the difference between RHS and LHS.
Let's claim:
$$b^{a+1} - c^{a+1} + (a+1)^c - (a+1)^b > b^a - c^a + a^c - a^b, \text{ for all } a \ge b > c \ge 2.$$
Rewrite the inequality:
$$(b-1)b^a - (c-1)c^a + (a+1)^c - a^c > (a+1)^b - b^b \quad (1)$$
Notice that $(a+1)^c - a^c > 0$ and $(a+1)^b < (1+\frac{1}{a})^b \cdot a^b < 3a^b$. Thus, if the following:
$$(b-1)b^a - (c-1)c^a > 2a^b \quad (2)$$
holds, then so does the claim. By lemma 1, we have $(\frac{b}{c})^a \ge (1+\frac{1}{c})^c > 2$. So
$$(b-1)b^a - (c-1)c^a > \left(b-1-\frac{c-1}{2}\right)b^a \ge \left(\frac{c+1}{2}\right)b^a.$$
If $c \ge 3$, $(\frac{c+1}{2}) b^a \ge 2b^a > 2a^b$. Also, when $c=2, b \ge 4$, Eq. (2) holds by similar argument.
It remains to check for $b=3, c=2$ for Eq.(1), that is,
$$2 \cdot 3^a - 2^a + (a+1)^2 - a^2 > (a+1)^3 - a^3$$
which clearly holds. Finally, note that when $a=b$, it's an equation, therefore, the inequality holds.

Case 1. $a > b > c$: By lemma 3, $LHS < RHS$. In particular, there is no solution.

Case 2. $b > a > c$: By lemma 3 again, $RHS < LHS$. Still no solution. So the possibilities are permutations of $(x, y, y)$ where $x, y \in N$ with $\min\{x, y\} = 2$, which indeed satisfy the equation. Next, when $c = 1$, it becomes
$$a^b - b^a = a - b$$
WLOG $a > b$, then
$$a^b - b^a = a - b > 0 \rightarrow a^b > b^a.$$
By lemma 2, we must have $b < 3$. In other words, $b = 1$ or $2$. If $b = 1$, then $a$ can be arbitrary positive integer. If $b = 2$, then
$$2^a = a^2 - a + 2 = (a + 1)(a - 2)$$
which means $a + 1, a - 2$ are powers of 2. Thus, they must be $4, 1$, respectively, since the difference between distinct powers of 2 is at least 3. The equality holds if and only if they are $1, 4$. So, $a = 3$ and we conclude the solutions are permutations of $(1, 2, 3)$ and $(x, y, y)$ with $x, y \in N$.