Singapore Mathematical Olympiad (SMO)

We first check that $n=1,2,4$ are solutions among $n\le 4$, with $m=1,2,5$ respectively.

We now assume that $n\ge 5$. By Legendre's formula, we know that $v_2(n!)=n-s_2(n)$ where $s_2(n)$ is the number of non-zero digits in the binary representation of $n$. Thus $v_2(n!)\le n-1$.

If $v_2(n!)=a<n-1$, letting odd($x$) denote the odd part of $x$, we have $$\mathrm{odd}(n!)+2^{n-1-a}=2^{m-a}$$ which cannot happen as $\mathrm{odd}(n!)$ is odd and the other two terms are even. Hence $v_2(n!)=n-1$ and we must have $s_2(n)=1\Rightarrow n$ is a power of 2.

Let $n=2^a$. Then $a\ge 3$. We claim that $\mathrm{odd}(n!)\equiv 3(\mathrm{mod}8)$. To see so, we pair up $i$ with $n-i=2^a-i$. If $v_2(i)<a-2$, then $\mathrm{odd}(2^a-i)\equiv -\mathrm{odd}(i)(\mathrm{mod}8)$ and so we have $$\mathrm{odd}(i)\cdot \mathrm{odd}(2^a-i)\equiv -\mathrm{odd}(i{)}^2\equiv -1(\mathrm{mod}8)$$ and we have an even number of such pairs. This gives us a product of 1. The remaining $i$'s are of simply $2^{a-2},2^{a-1}$ and $3\cdot 2^{a-2}$, whose odd part multiply to 3 mod 8. Hence the total product is 3 mod 8 as desired. Then $$n!+2^{n-1}=2^m\Rightarrow \mathrm{odd}(n!)+1=2^{m-n-1}\Rightarrow 2^{m-n-1}\equiv 4(\mathrm{mod}8)\Rightarrow 2^{m-n-1}\le 4$$ Thus $\mathrm{odd}(n!)\le 4$ which leads to $n\le 4$. So there are no other solutions.