Example 1 Let $n$ be a given positive integer, find the number of positive integer solutions $(x, y)$ for
$$
\frac{1}{n}=\frac{1}{x}+\frac{1}{y} \quad(x \neq y)
$$
From (1), we know that the positive integer solutions $(x, y)$ satisfy $x>n, y>n$. We can let
$$
x=n+r, y=n+s \quad(r \neq s) .
$$

Substituting (2) into (1) yields
$$
\frac{1}{n}=\frac{1}{n+r}+\frac{1}{n+s}=\frac{2 n+r+s}{(n+r)(n+s)} .
$$

Thus,
$$
n^{2}+(r+s) n+r s=2 n^{2}+(r+s) n \text {. }
$$

We get
$$
n^{2}=r s .
$$
The different divisors $r$ of $n^{2}$ are $d\left(n^{2}\right)$ in total, but we need to exclude the case $r=n$. Therefore, 
the number of positive integer solutions $(x, y)$ for (1) is $d\left(n^{2}\right)-1$.