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 $\mathrm{odd}(n!) \equiv 3 \pmod 8$. To see so, we pair up $i$ with $n-i = 2^a-i$. If $v_2(i) < a-2$, then $\mathrm{odd}(2^a-i) \equiv -\mathrm{odd}(i) \pmod 8$ and so we have
$$\mathrm{odd}(i) \cdot \mathrm{odd}(2^a - i) \equiv -\mathrm{odd}(i)^2 \equiv -1 \pmod 8$$
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
$$n! + 2^{n-1} = 2^m \Rightarrow \mathrm{odd}(n!) + 1 = 2^{m-n-1} \Rightarrow 2^{m-n-1} \equiv 4 \pmod 8 \Rightarrow 2^{m-n-1} \le 4$$
Thus $\mathrm{odd}(n!) \le 4$ which leads to $n \le 4$. So there are no other solutions.