Answer: $(1, 1)$, $(2, 1)$, $(5, 3)$.

By checking trivial cases $m = 1$, $n = 1$ and $n = 2$, we find the first two of the solutions listed. From now on, we assume $m \ge 2$ and $n \ge 3$.

Step 1. We prove that $m > n$.
Suppose $n \ge m$. Then, numbers $m!$, $n!$ and $m^n$ are divisible by $m$, hence, $m$ also divides $m! + n! - m^n = 1$ – contradiction. In particular, it follows that $m \ge 4$.

Step 2. Now, we show that $n > \frac{1}{2}m$.
From well-known inequality $m! > m^{\frac{1}{2}m}$, which follows from the Root-Mean Square-Arithmetic-Mean-Geometric-Mean-Harmonic-Mean Inequality, we obtain $m^n + 1 = m! + n! > m^{\frac{1}{2}n} + 1$.

Step 3. We prove that $m$ is prime.
Otherwise, $m$ has a prime factor $p$ not larger than $\frac{1}{2}m$ and $n$. In that case, $p$ is a factor of $m! + n! - m^n = 1$ – contradiction.

Step 4. We prove that $m-1$ is either prime or a square of a prime. Suppose the contrary. Then, $m-1$ can be represented as a product of two factors $a \ne b$, both smaller than $\frac{1}{2}m$ and $n$. Therefore, $m!$ and $n!$ are divisible by $n-1 = ab$, since $n!$ contains both these factors. $m^n + 1 = ((m-1) + 1)^n + 1$ gives a remainder of 2 when divided by $m-1$. Hence, $m-1$ is a factor of 2, which is impossible, since $m \ge 4$.

From the results of Steps 3 and 4 and from parity, we get that the only possible case is $m = 5$. Indeed, since $m \ge 4$ is prime, it is odd, hence, $m-1$ is even and can be a square of a prime, if that prime is 2. From the results of Steps 1 and 2, we get that possible solutions are only $(5, 3)$ and $(5, 4)$. By checking, we see that only the first pair satisfies the initial equation.