Example 17 (1991 Japan Mathematical Olympiad Preliminary Problem) How many pairs of positive integer solutions $(x, y)$ satisfy $\frac{1}{x+1}+\frac{1}{y}+\frac{1}{(x+1) y}=\frac{1}{1991}$?
Solution: Clearly, $x+1, y$ are both greater than 1991, so the original equation can be transformed into
$$
(x+1-1991)(y-1991)=1991 \times 1992=2^{3} \cdot 3 \cdot 11 \cdot 83 \cdot 181 .
$$
$x+1-1991$ can be any divisor on the left side of the equation, so the number of $x$ is equal to the number of positive divisors of $1991 \times 1992$, which is $(3+1)(1+1)(1+1)(1+1)(1+1)=2^{6}=64$. Therefore, there are 64 pairs of positive integer solutions $(x, y)$ that satisfy the condition.