The solutions are $n = 1, k = 1$; $n = 2, k = 1$; $n = 4, k = 2$.
It is easy to check that the solutions above are the only solutions for $n \le 4$. So assume $n > 4$. So $n! + 1 > n + 1$, so $k > 1$. If $n$ is odd, then $n + 1$ is even, but $n! + 1$ is odd, so there are no solutions. So $n$ is even. Hence $n$ is composite. So $n$ divides $(n-1)!$. But using the binomial theorem we have $(n+1)^k - 1 = n^k + \binom{k}{1}n^{k-1} + \cdots + \binom{k}{k-2}n^2 + kn$. Hence $(n-1)! = n(n^{k-2} + n^{k-3} + \cdots + \binom{k}{k-2}) + k$. Hence $n$ divides $k$. But that means $k \ge n$, and $(n+1)^n > n! + 1$. So there are no other solutions.