It is clear that if $n \ge 3$, $n!$ is divisible by $3$, that is $n! = 3k$, for some natural number $k$. But then $2^{n!} = 2^{3k} = 8^k \equiv 1 \pmod{7}$. It's easy to see that the cubes of integer numbers are equal to $0$ or $\pm 1$ modulo $7$. Without loss of generality, we may assume that $a \ge b$.

Case 1. $a \ge 3$. If $b \ge 3$, then $2^{a!} + 2^{b!} \equiv 2 \pmod{7}$ is not a cube of an integer number. If $b=2$, then $2^{a!} + 2^{b!} \equiv 1+4 \equiv 5 \pmod{7}$ is not a cube of an integer number. If $b=1$, then $2^{a!} + 2^{b!} \equiv 1+2 \equiv 3 \pmod{7}$ is not a cube of an integer number.

Case 2. $2 \ge a \ge b$, here we have three cases, from which we simply find a single answer in a simple overview.