Example 12 (1991 Japan Mathematical Olympiad) Find the positive integer solutions of the indeterminate equation $\frac{1}{x+1}+\frac{1}{y}+\frac{1}{(x+1) y}=$ $\frac{1}{1991}$.
Given that the equation can be transformed into
$$
\begin{array}{l}
1991 y + 1991(x + 1) + 1991 = (x + 1) y, \\
(x + 1) y - 1991(x + 1) - 1991 y = 1991, \\
(x + 1) y - 1991(x + 1) - 1991 y + 1991^2 = 1991 + 1991^2, \\
{[(x + 1) - 1991](y - 1991) = 1991 \cdot 1992 .}
\end{array}
$$

Obviously, the number of positive integer solutions to the given equation is equivalent to the number of positive divisors of \(1991 \cdot 1992\). Since
$$
1991 \cdot 1992 = 2^3 \cdot 3 \cdot 11 \cdot 83 \cdot 181 \text{, }
$$

\(1991 \cdot 1992\) has
$$
(3+1)(1+1)(1+1)(1+1)(1+1) = 64
$$

positive divisors, meaning the given equation has 64 sets of positive integer solutions.