*Answer.* The only solutions are $(1, 1, 1)$ and $(2, 2, 2)$.

We can assume without loss of generality that $a \le b$.

* For $a = b = 1$, we get $c = 1$, which gives the solution $(1, 1, 1)$.

* For $a = 1$ and $b > 1$, the left-hand side is bigger than $1$ and odd, therefore, it cannot be a power of $2$ and we do not get a solution in this case.

* For $a = b = 2$, we get $c = 2$, therefore $(2, 2, 2)$ is a solution.

* For $a = 2$ and $b = 3$, we get $2! + 3! = 8 = 2^3$. But there is no $c$ with $c! = 3$. Therefore, there is no solution in this case.

* For $a = 2$ and $b \ge 4$, we get $2! + b! \ge 2! + 4! = 26$. Therefore, we have $c! > 4$. This implies that the left-hand side is congruent to $2$ modulo $4$, while the right-hand side is congruent to $0$ modulo $4$. Therefore, there is no solution in this case.

* For $a \ge 3$ and $b \ge 3$, the left-hand side is divisible by $3$ while the power of $2$ on the right-hand side is not. Therefore, there is no solution in this case.

(Reinhard Razen) ☐