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$m\ge5$. 1 Let us look for a probabilistic hint. Given$n\ge3$, define$N:=[n!]$, the integer part of$n!$. 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\$.