Let $p$ be a prime dividing $m$. We then have $x^n-1=(x-1)(x-2)\ldots(x-n)$ in $\mathbb Z_p[x]$. We then have $n<p$ and $1+2+\ldots+n=0\pmod p\iff p|\frac{n(n+1)}2\iff n=p-1$. This means that $m$ is a prime power, $p^t$, and $n$ is $p-1$. However, it's well-known that there is a primitive root $\pmod{p^2}$ in the set $\{1,2,\ldots,p-1\}$ (it's even been discussed on the forum), so unles $t=1$, this can't be possible, meaning that the pairs are $(m,n)=(p,p-1)$ with prime $p$ (it's easy to check that these pairs work).

Thaks grobber, I did the same thing you did. Another way to prove the last part is to notice that: $(p-1)^2-1=p^2-2p=0(p)$, and different from $0$ $mod p^2$ (p=2 is dealed manually). so, $(p-1)^{2\left(\frac{p-1}{2}\right)}=\left((p-1)^2-1\right)\times A$, where $(A,p)=1$.

Let $p$ be a prime dividing $m$. We then have $x^n-1=(x-1)(x-2)\ldots(x-n)$ in $\mathbb Z_p[x]$ so $1+2+\ldots+n=0\pmod p\iff p|\frac{n(n+1)}2\iff n=p-1$.($n=p+1$ is impossible) A similar argument shows that $m=p^k$ so to show this we will show that $(p-1)^{p-1}+1 \equiv (-1)^{p-1}+(-1)^{p-2}(p-1)^2 \equiv -(p-1)^2+1 \neq 0 \pmod {p^2}$ so the only solutions are $(m,n)=(p,p-1)$ @below,Reduce all the coefficients of $F(x):=\prod_{d=1}^n (x-d) \pmod p$.If $F \not\equiv X^n-1$ in $[\mathbb{F}_p]$ then subtracting them gives a polynomial of degree $n-1$ having $n$ zeroes modulo $p$ contradiction by Lagrange's Theorem.So all the coefficients of $F$ except the coefficient of $X^n$ and the constant term is divisible by $p$.Vieta's theorem finishes.

still couldn't solve it after 2 hours:D