5. Find all positive integers $a, b$ such that $2^{a!}+2^{b!}$ is a cube of some positive integer.
5. $a=2, b=2$.

Obviously, when $n \geqslant 3$, we have $n!\equiv 0(\bmod 3)$.
Thus, there exists a positive integer $k$ such that $n!=3 k$.
Therefore, $2^{n!}=2^{3 k}=8^{k} \equiv 1(\bmod 7)$, and the cube of a positive integer modulo 7 can only be $0, \pm 1$.
Without loss of generality, assume $a \geqslant b$.
(1) $a \geqslant 3$.
If $b \geqslant 3$, then
$2^{a!}+2^{b!} \equiv 2(\bmod 7)$,

which cannot be a cube number.
Similarly,
when $b=2$, $2^{a!}+2^{b!} \equiv 5(\bmod 7)$;
when $b=1$, $2^{a!}+2^{b!} \equiv 3(\bmod 7)$,

neither of which can be a cube number.
(2) $2 \geqslant a \geqslant b$.
By calculating each case, we find that only $a=2, b=2$ satisfies the condition.