Example 5 Find the integer solutions of the indeterminate equation $x^{2}(y-1)+y^{2}(x-1)=1$.

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Let $(x, y)$ be an integer solution of the equation, and assume $x \leqslant y$, then $y \geqslant z$. At this point, consider the original equation as a quadratic equation in $x$: $(y-1) x^{2}+y^{2} \cdot x-\left(y^{2}+1\right)=0$.
Since equation (1) has integer solutions, $\Delta=y^{4}+4(y-1)\left(y^{2}+1\right)$ must be a perfect square, i.e.,
$$
\left(y^{2}+2 y\right)^{2}-8 y^{2}+4 y-4
$$

is a perfect square.
Notice that when $y \geqslant 8$, $\left(y^{2}+2 y-3\right)^{2}>\Delta>\left(y^{2}+2 y-4\right)^{2}$, and when $2\Delta>\left(y^{2}+2 y-3\right)^{2}$, thus $y>2$, $\Delta$ is not a perfect square. Therefore, $y=2$, and then $x=1$ or -5.
In summary, the integer solutions of the equation are $(x, y)=(1,2),(2,1),(-5,2)$ or $(2,-5)$.