8,9
Find all natural $m$ and $n$, for which $m!+12=n^{2}$.
When $m \geq 5$, $m!$ is even and a multiple of 5, meaning the decimal representation of this number ends in zero, and the number $m!+12$ ends in the digit 2. Therefore, it cannot be a perfect square.

By checking the values $m=1,2,3,4$, we get that $m=4, n=6$.

## Answer

$m=4, n=6$.