Example 7 Given that $n$ is a positive integer, and $n^{2}-71$ is divisible by $7 n+55$. Try to find the value of $n$.
(1994-1995, Chongqing and Four Other Cities Mathematics Competition)
Solution: Let $\frac{n^{2}-71}{7 n+55}=k$ (where $k$ is an integer),
then $n^{2}-7 k n-(55 k+71)=0$.
$$
\begin{aligned}
\Delta & =49 k^{2}+4(55 k+71) \\
& =49 k^{2}+220 k+284 .
\end{aligned}
$$

When $\Delta$ is a perfect square, the $n$ in equation (1) is an integer. And
$$
\begin{array}{l}
(7 k+15)^{2}<\Delta=49 k^{2}+220 k+284 \\
<(7 k+17)^{2} . \\
\therefore \Delta=(7 k+16)^{2} .
\end{array}
$$

Thus, $(7 k+16)^{2}=49 k^{2}+220 k+284$.
Solving this, we get $k=7$.
Substituting $k=7$ into (1), we get $n=57$. (discard -8)