$m = 1$ gives no solution, therefore $m!$ is even and $n$ should also be even. Let $n = 2^t \cdot s$ ($t$ and $s$ are positive integers, $t \ge 1$ and $s$ is odd). $t = 1$ readily leads to $n = 2$ and $m = 3$.

Now let $t \ge 2$. Then $m! = 2^n + n = 2^{2t \cdot s} + 2^t \cdot s \ge 2^{2t} + 2^t$. By induction over $t$ we will show that
$$2^{2t} + 2^t > (2t - 1)! \quad (1)$$
For $t=2$, $3$ the inequality holds. Assume that it is held for $t=k$. In order to show that it also holds for $k+1$ we have to prove that
$$\frac{2^{2k+1} + 2^{k+1}}{2^2 + 2^k} \ge 2k(2k+1)$$
Since $\frac{2^{2n} + 2n}{2^n + n} \ge 2^{n-1} \iff 2^{2n-2} \ge n(2^{n-2} - 1)$ and $2^{2n-2} \ge n \cdot 2^{n-2}$ for each positive integer $n$ we get $\frac{2^{2n} + 2n}{2^n + n} \ge 2^{n-1}$. By taking $n = 2^k$ we get $\frac{2^{2^{k+1}} + 2^{k+1}}{2^2 + 2^k} \ge 2^{2^{k-1}}$. Thus, in order to complete the proof we will show that $2^{2^{k-1}} \ge 2k(2k+1)$ for $k \ge 3$. For $k = 3, 4$ it is held since $2^7 > 42$ and $2^{15} > 72$ and for $k \ge 5$ we have $2^{2^{k-1}} \ge 2^{4k} = 2^{2k} \cdot 2^{2k} \ge 2k(2k+1)$. (1) is proved.

By (1) $m! > (2t-1)!$ and consequently $m \ge 2t$. Since $2^n + n = 2^{2t \cdot s} + 2^t \cdot s = 2^t(2^{2t \cdot s-t} + s)$ and $2^t \cdot s \ge 2^t > t$ we get that $2^{2t \cdot s-t} + \beta$ is odd. Therefore $2^t$ divides the maximal even factor of $m!$ and since $m \ge 2t$ we get
$$\alpha = \left\lfloor \frac{m}{2} \right\rfloor + \left\lfloor \frac{m}{4} \right\rfloor + \dots \ge t + \left\lfloor \frac{t}{2} \right\rfloor + \left\lfloor \frac{t}{4} \right\rfloor + \dots$$
Contradiction since $t \ge 2$. Thus, the only solution is: $(m, n) = (3, 2)$.