Note that if $a \geq 3$ and $b \geq 3$, then $3 | 2^{c!}$ which is a contradiction. So we just check the four possibilities $(1, 1), (2, 1), (2, 2), (1, 2)$. WLOG, $a = 1$. $b = 1$ or fails by parity. Now consider, $a = 2$, then, for $b \geq 4$, $c \geq 3$ does not work. $ c =2$ works when $b=2$. $c= 1$ works only when $b = 1$. Check all possible. Get $(a,b) = (1, 1, 1,), (2, 2, 2)$.

LogiBobb wrote: Note that if $a \geq 3$ and $b \geq 3$, then $3 | 2^{c!}$ which is a contradiction. So we just check the four possibilities $(1, 1), (2, 1), (2, 2), (1, 2)$. WLOG, $a = 1$. $b = 1$ or fails by parity. Now consider, $a = 2$, then, for $b \geq 4$, $c \geq 3$ does not work. $ c =2$ works when $b=2$. $c= 1$ works only when $b = 1$. Check all possible. Get $(a,b) = (1, 1, 1,), (2, 2, 2)$. What if a>= 3 but b< 3?

We can check the answers $(a,b,c) = (1,1,1)$, now let $c \ge 2$ Note, that if both $a$ and $b$ are $\ge 3$, then $2^c$ won't divide $3$, contradiction. So, one of $a$ or $b$ is $\leq 2$. W.L.O.G. $a \ge b, b \leq 3$. We know that $b$ won't divide 4, and If $a \ge 4$, then $a! + 2 \equiv 0 + 2 \equiv 2 (mod 4$, but $2^{c!} \equiv 0$, contradcition. Then just check all the possibilites for $a,b$ and we get the 2nd answer $(a,b,c)=(2,2,2)$