Let us look for a probabilistic hint. Given $n\ge3$, define $N:=[n!]$, N:=[\sqrt{n!}]$, the integer part of the square root $n!$ . Then $n!\in[N^2+1,\ldots,N^2+2N]$. The answer to the question is positive if and only if $n!=N^2+N$, because $m$ has to be $N$. At first glance, the probability of this event is $1/2N$. However, we know a priori that both $n!$ and $N(N+1)$ are even. Therefore this probability is $1/N\sim(n!)^{-1/2}$.
Since the series $$\sum_{n=2}^{\infty}\frac{1}{\sqrt{n!}}$$ converges, I expect that the number of solutions to this problem be finite. Actually, I checked that the answer is No for $4\le n\le 10$. Then the number of solutions with $n\ge4$ can be estimated by the series $$\sum_{n=11}^{\infty}\frac{1}{\sqrt{n!}}$$ Because this number is very small (not greater than $10^{-3}$), I bet that there does not exist a solution $n\ge4$.
This is the same kind of reasoning that is used to guess that there does not exist a prime number among Fermat numbers $F_m$ with $m\ge5$.
