Without loss of generality, assume that $m \ge n$. Then, if $n > 1$ equation can be written the following way:
$$1 \cdot 2 \cdot 3 \cdots m \cdot (m+1) \cdot (m+2) \cdots (m+n) = 2 \cdot 1 \cdot 2 \cdot 3 \cdots m \cdot 1 \cdot 2 \cdot 3 \cdots n \Rightarrow \\ (m+1) \cdot (m+2) \cdots (m+n) = 2 \cdot 1 \cdot 2 \cdot 3 \cdots n.$$
The factors on the left side are not less than the factors on the right side, since the factors can be paired the following way:
$$m+1 > 1,\ m+2 > 2,\ \dots,\ m+n-1 > n-1,\ m+n \ge 2n.$$
Since the factor $m+1$ always exists and $m+1 > 1$, equation doesn't hold. If $n=1$, equation $(m+1)! = 2m!$ holds only if $m+1=2$. Thus, the only solution is $m=n=1$.