Example 2 Find all integer pairs $(a, b)$, satisfying
$$
\left(b^{2}+11(a-b)\right)^{2}=a^{3} b .
$$
Notice,
$$
\begin{array}{l}
\left(b^{2}+11(a-b)\right)^{2}=a^{3} b \\
\Leftrightarrow b^{4}+22 b^{2}(a-b)+121(a-b)^{2}=a^{3} b \\
\Leftrightarrow b\left(a^{3}-b^{3}\right)-22 b^{2}(a-b)-121(a-b)^{2}=0 \\
\Leftrightarrow(a-b)\left(a^{2} b+a b^{2}+b^{3}-22 b^{2}-121(a-b)\right) \\
\quad=0 .
\end{array}
$$

If $a=b$, then we get infinitely many solutions $(a, a)$.
Next, consider the case $a \neq b$. In this case,
$$
a^{2} b+a b^{2}+b^{3}-22 b^{2}-121(a-b)=0 \text {. }
$$

If $b=0$, then $a=0$, which is a contradiction.
Thus, $b \neq 0$.
Rearranging equation (1) as a quadratic equation in $a$, we get
$$
b a^{2}+\left(b^{2}-121\right) a+b^{3}-22 b^{2}+121 b=0 \text {. }
$$

The discriminant is
$$
\begin{array}{l}
\Delta=\left(b^{2}-121\right)^{2}-4 b\left(b^{3}-22 b^{2}+121 b\right) \\
=(3 b+11)(11-b)^{3} .
\end{array}
$$

By $\Delta \geqslant 0 \Rightarrow-\frac{11}{3} \leqslant b \leqslant 11$.
Substituting $b=-3,-2,-1,1,2, \cdots, 11$ into equation (2), we find that except for $b=11$, $\Delta$ is not a perfect square.
When $b=11$, $a=0$, which satisfies the condition.
Therefore, the solution set is
$$
\{(t, t) \mid t \in \mathbf{Z}\} \cup\{(0,11)\} .
$$