The answer is $(3, 4)$.

For $m = 0, 1$ or $2$, we obtain the impossible $2n! = 3$ and $2n! = 8$. However, $m = 3$ works with $n = 4$.

Now let $m \ge 4$. Clearly, $2n! > 2m!$, which implies $n > m$, so we can write $n = m + a$ for some positive integer $a$. The equation transforms into
$$2(m + 1)(m + 2)(m + a) = m! + 2,$$
and further simplifies to
$$(m+1)(m+2)(m+a) = \frac{m!}{2} + 1.$$
For $m \ge 4$, the right-hand side is odd, but for $a \ge 2$, the left-hand side is even.
Therefore, we have no solutions for $m \ge 4$ and $a \ge 2$.

Finally, if $n = m+1$, then we have $2(m+1) = m!+2$, which simplifies to $(m-1)! = 2$.
This is impossible for $m \ge 4$.