Example 7 Find all integer solutions to the indeterminate equation $x y=x^{2}+6$.

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Solve: From $x y=x^{2}+6$ we get $x(x-y)=-6$, so $x \mid(-6)$. When $x=1$, $y=7$; when $x=-1$, $y=-7$. When $x=2$, $y=5$; when $x=-2$, $y=-5$. When $x=3$, $y=5$; when $x=-3$, $y=-5$. When $x=6$, $y=7$; when $x=-6$, $y=-7$. Therefore, all integer solutions to $x y=x^{2}+6$ are
$$\left.\left.\left.\left.\left.\left.\begin{array}{l}
x=1 \\
y=7
\end{array}\right\}, \quad \begin{array}{l}
x=-1 \\
y=-7
\end{array}\right\}, \quad \begin{array}{l}
x=2 \\
y=5
\end{array}\right\}, \quad \begin{array}{l}
x=-2 \\
y=-5
\end{array}\right\}, \quad \begin{array}{l}
x=3 \\
y=5
\end{array}\right\}, \quad \begin{array}{l}
x=-3 \\
y=-5
\end{array}\right\}, \quad \begin{array}{l}
x=6 \\
y=7
\end{array}\right\}, \quad \begin{array}{l}
x=-6 \\
y=-7
\end{array}\right\}$$