First check that $(m,n)=(1,1),(2,1),$ and $(5,3)$ are the only solutions with $m\le 5.$ We will now show that there are no solutions for $m,n \ge 5.$ Consider the equation $m!+n!=m^n+1 \text{ (*)}$ in $\bmod 3.$ We have $m^n \equiv 2 \bmod 3 \implies m \equiv 2 \bmod 3, n \equiv 1 \bmod 2\text{ (**)}.$ Then consider $(*)$ $\bmod 5:$ $m^n \equiv 4 \bmod 5 \implies m \equiv 2,3,4 \bmod 5,$ with $n \equiv 1 \bmod 4, n \equiv 3 \bmod 4,$ and $n \equiv 0 \bmod 2,$ respectively $(***)$ (all of this makes sense since $m,n \ge 5$). However, the last case contradicts $(**),$ so in reality there are only two possibilities when $(**)$ is combined with $(***):$ $m \equiv 2 \bmod 15,n \equiv 1 \bmod 4$ and $m \equiv 8 \bmod 15,n \equiv 3 \bmod 4.$ Now, note that $m$ must be odd, so we have a tighter restriction of $m \equiv 17 \bmod 30,n \equiv 1 \bmod 4$ $(*****)$ and $m \equiv 23 \bmod 30,n \equiv 3 \bmod 4$ $(******).$ Plugging $(*****)$ into $(*)$ and taking the entire equation $\bmod 30$ yields (for some integer $k \ge 1$) $17^{4k+1}\equiv -1 \equiv 119 \bmod 30 \implies 17^{4k}\equiv 7 \bmod 30.$ But for any number $a$ relatively prime to $30,$ $a^{4k} \equiv 1 \not\equiv 7 \bmod 30,$ which leads to a contradiction. We get a similar contradiction if we plug $(******)$ into $(*),$ thus proving that $n$ and $m$ cannot be bigger than or equal to $5$ simultaneously and showing that the only solutions are $(m,n)=(1,1),(2,1),$ and $(5,3).$