Solve $n^3-m!=40$, $m$ and $n$ are positive integers.

Question

Solve $n^3-m!=40$, $m$ and $n$ are integers. $m=n=4$ is an obvious answer. I could not come to find any other solution. Is there any?

Answer 1

If $m\ge10$, then $5\mid40+m!$ but $5^3\not\mid40+m!$, so it suffices to check for cubes with $m\le9$.

Added later: Alternatively, if $m\ge6$, then $40+m!\equiv4$ mod $9$, but the only cubes mod $9$ are $0$, $1$, and $8$, so it suffices to check for cubes with $m\le5$.