The moment I saw I was motivated to approach this problem by analyzing the powers of two. In this solution, define the maximum power of that divides .

Since , hence because if not, let , then divide both sides by $min({2^{q(m!)},2^{n})$ to get a contradiction, the same thing can be done for . Note that this condition means .(This suggests a finite number of solution and we can prove that using inequality).

It's convenient to make a list of values for . Testing starting with small values gives . At it's pretty clear that there's no solution for larger values.

We claim that for .

To prove this we have to find a lower bound for in terms of so that the inequality is more manageable. One well-known way to calculate is .

We can check by hand that our claim is true for .

For .

We now proceed by proving . We take on both sides: . Since (note that ), we increase our bound: which is true for our domain.

Hence $2^{2^{q(m!)}}+2^{q(m!)}\ge 2^{2^{\frac {m}{2}+2}}+2^{\frac {m}{2}+2}}>m!$ for all , and this means our claim has been proven.

In conclusion is the only solution.