59th Ukrainian National Mathematical Olympiad

It is clear that if $n\ge 3$, $n!$ is divisible by $3$, that is $n!=3k$, for some natural number $k$. But then $2^{n!}=2^{3k}=8^k\equiv 1(\mathrm{mod}7)$. 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 $2^{a!}+2^{b!}\equiv 2(\mathrm{mod}7)$ is not a cube of an integer number. If $b=2$, then $2^{a!}+2^{b!}\equiv 1+4\equiv 5(\mathrm{mod}7)$ is not a cube of an integer number. If $b=1$, then $2^{a!}+2^{b!}\equiv 1+2\equiv 3(\mathrm{mod}7)$ 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.