Determine all pairs of positive integers $(m,n)$ that satisfies $m!+n!=m^n$ Determine all pairs of positive integers $(m,n)$ that satisfies $m!+n!=m^n$
I found easily the pairs $(2,2)$ and $(2,3)$ but i can't prove that these are the only pairs possibles.
Any hints?

This is not quite a full answer but it has the main ideas you'll need.
First, let us get the ability to assert a lower bound on $m$ after checking only finitely many pairs. This can be done by noting that $m^n>n!$, so by Stirling's approximation, $n<em$. Thus we can divide the problem between $m<m^*$ and $m \geq m^*$ by directly checking the cases $m=1,2,\dots,m^*-1$ and $n=1,2,\dots,\lfloor em \rfloor$.
Second, let us see what happens if $m$ is large. Similar to the above, if we assume $m>e$, then Stirling's approximation applied to $m^n>m!$ implies 
$$m \leq \frac{n}{1-1/\ln(m)}.$$
So it suffices to provide an upper bound on $n$ in order to finish this case.
Third, let $r=\min \{ m,n \}$. Then $r!$ divides $m^n$ and so $r\#$ divides $m$, where $\#$ is the primorial function. Thus in particular $r\# \leq m$. 
So there are two cases: $r=m$ and $r=n$. Consider $r=m$. By Bertand's postulate, if $m>2$ then there is a prime $p$ between $m/2$ and $m$, and so $m\# \geq 2p>m$. So $r=m$ can only occur for $m \leq 2$.
Now consider $r=n$. In this case $n\# \leq m$. If we assume $m \geq 8$, then $1-1/\ln(m) \leq 2$, so for $m \geq 8$ the inequality above gives
$$n\# \leq 2n.$$
Applying Bertrand's postulate twice in the same manner as the previous case implies that $n\# \geq n^2/8$. Thus $n \leq 16$.
Putting the pieces together, we divide the problem between $m<8$ and $m \geq 8$. In the case $m<8$ we have $n \leq \lfloor em \rfloor$, which is on the order of 100 cases to check. In the case $m \geq 8$ we have $n \leq 16$ and $m \leq 2n$, which is again on the order of 100 cases to check.
