11. (20 points) Find all positive integer pairs \((n, k)\) that satisfy
$$
(n+1)^{k}-(2 n+1)=n!
$$
11. Notice,
$$
\begin{array}{l}
(n+1)^{k}-(2 n+1)=n! \\
\Leftrightarrow(n+1)\left[(n+1)^{k-1}-1\right]=n[(n-1)!+1] .
\end{array}
$$

Let $p$ be a prime factor of $n+1$.
Since $(n+1, n)=1$, we know $p \mid (n-1)!+1$.
For any prime $q (q \leqslant n-1)$, we have $q \mid (n-1)!+1$.
Thus, $p=n+1$.
$$
\text { Equation (1) } \Leftrightarrow p^{k}-(2 p-1)=(p-1)!\text {. }
$$

Therefore, $(p-1)!\equiv 1(\bmod p)$.
When $p=2$, Equation (2) becomes $2^{k}=4$, solving for
$$
k=2, n=1 \text {. }
$$

When $p$ is an odd prime, construct a polynomial of degree $p-2$
$$
f(x)=\prod_{i=1}^{p-1}(x-i)-\left(x^{p-1}-1\right) \text {. }
$$

For $x=1,2, \cdots, p-1$, we have $p \mid \prod_{i=1}^{p-1}(x-i)$.
By Fermat's Little Theorem, we know $p \mid\left(x^{p-1}-1\right)$.
Thus, $p \mid f(x) (x=1,2, \cdots, p-1)$
$$
\begin{array}{l}
\Rightarrow f(x) \equiv 0(\bmod p)(\text { for any } x \in \mathbf{Z}) \\
\Rightarrow f(0)=(-1)^{p-1}(p-1)!+1 \equiv 0(\bmod p) \\
\Rightarrow(p-1)!\equiv-1(\bmod p),
\end{array}
$$

Combining with Equation (3), we get $p=2$, a contradiction.
$$
\text { Therefore, }(n, k)=(1,2) \text {. }
$$