Example 3 Find the integer solutions of the equation $x^{2}(y-1)+y^{2}(x-1)=1$.

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Solution: Let $(x, y)$ be the integer solutions. When $y=1$, $x=2$; when $x=2$, $y=1$ or $-5$;

When $y>1$, the equation $(y-1) x^{2}+y^{2} x-\left(y^{2}+\right.$ $1)=0$ has integer solutions, then
$$
\begin{array}{l}
\Delta=y^{4}+4\left(y^{2}+1\right)(y-1)=y^{4}+4 y^{3}-4 y^{2} \\
+4 y-4=\left(y^{2}+2 y\right)^{2}-8 y^{2}+4 y-4 \\
=\left(y^{2}+2 y-2\right)^{2}-\left(4 y^{2}-12 y+8\right) \\
=\left(y^{2}+2 y-3\right)^{2}-\left(2 y^{2}-16 y+13\right) \\
=\left(y^{2}+2 y-4\right)^{2}+(20 y-20), \\
\because 4 y^{2}-12 y+8>0 \Leftrightarrow y>2,2 y^{2}-16 y+ \\
13>0 \Leftrightarrow y \geqslant 8,20 y-20>0 \Leftrightarrow y>1,
\end{array}
$$
$\therefore$ When $y \geqslant 8$, $\left(y^{2}+2 y-4\right)^{2}2$ is not a perfect square, while when $y=2$, $x^{2}+4 x-5=0$ gives $x=1,-5$.

Therefore, the solutions to the original equation are $(x, y)=(2,1),(1,2)$, $(-5,2),(2,-5)$.