11. (2007 National Team Training Test Question) Find all positive integer pairs $(a, b)$, such that: $a^{2}+b+1$ is a prime number, $a^{2}+b+1$ divides $b^{2}-a^{3}-1$, and $a^{2}+b+1$ does not divide $(a+b-1)^{2}$.
11. From the given, $\frac{(b+1)(a+b-1)}{a^{2}+b+1}=\frac{b^{2}-a^{3}-1}{a^{2}+b+1}+a$ is also an integer.

Let $p^{k}=m=a^{2}+b+1, k$ be a positive integer, then $p^{k} \mid (b+1)(a+b-1)$.
If $p$ does not divide $b+1, a+b-1$ one of them, then $p^{k}$ must divide the other, but in this case. $p^{k}=a^{2}+b+1>\max \{b+1, a+b-1\}$, a contradiction! Therefore,
$$
\begin{array}{l}
b+1 \equiv a+b-1 \equiv 0(\bmod p), \\
a^{2}=p^{k}-(b+1) \equiv 0(\bmod p) .
\end{array}
$$

Since $p$ is a prime, we have
$a \equiv 0(\bmod p)$,
$$
0 \equiv (b+1)-(a+b-1)+a \equiv 2(\bmod p) \Rightarrow p=2 \text{. }
$$

Let
$$
\left\{\begin{array}{l}
a^{2}+b+1=2^{k}, \\
b+1=2^{k_{1}} t_{1}, \\
a+b-1=2^{k_{2}} t_{2}, \\
k_{1}+k_{2} \geqslant k, \\
k>2 k_{2},
\end{array}\right.
$$

where $t_{1}, t_{2}$ are odd numbers, $k_{1}, k_{2}$ are positive integers, the last inequality is derived from $m \times (a+b-1)^{2}$.
Since $2^{k}=a^{2}+b+1>b+1=2^{k_{1}} t_{1}$, we also have
$k>k_{1}$.
(i) When $k_{1} \geqslant 3$,
Since $a^{2}=2^{k}-2^{k_{1}} t_{1} \equiv 0\left(\bmod 2^{3}\right)$, hence
$a \equiv 0(\bmod 4), a+b-1 \equiv 0-1-1 \equiv 2(\bmod 4)$.
Thus, $k_{2}=1$ must hold. From $k_{1}+k_{2} \geqslant k>k_{1}$, we know, $k=k_{1}+1$. At this time, $2^{k_{1}} t_{1}=b+1k_{1}$, we know, $k \geqslant 3$.
From $k_{1}+k_{2} \geqslant k$ and $k>2 k_{2}$, we have,
$2 k_{2}<k \leqslant k_{2}+2 \Rightarrow k_{2}<2$,
thus, $k \leqslant 3$, hence
$$
k=3 \text{. }
$$

But
$$
\left\{\begin{array}{l}
b+1=2^{2} t_{1}, \\
a^{2}+b+1=8
\end{array} \Rightarrow \left\{\begin{array}{l}
t_{1}=1, \\
a=2, \\
b=3,
\end{array}\right.\right.
$$

discard.
(iii) When $k_{1}=1$, then
$2 k_{2}<k \leqslant k_{2}+1 \Rightarrow k_{2}<1$, a contradiction!
In summary, the required tuple $(a, b)$ is $\left(2^{x}, 2^{2 x}-1\right)(x=2,3,4, \cdots)$.