(13) The positive integer solutions $(x, y, z)$ for the equation $2(x+1)(y+1)-1=x y z \quad(x<y)$ are $\qquad$ (list all possible cases).
Solution: $(1,3,5),(3,7,3)$.
$$
\begin{array}{l}
2 x y+2 x+2 y+1=x y z . \\
\therefore x y \mid (2 x y+2 x+2 y+1), \quad \therefore x y \mid (2 x+2 y+1), \quad x y \leqslant 2 x+2 y+1 .
\end{array}
$$

From $x2$, we know that $y$ does not exist.
If $x=3$, then $3 y \mid (2 y+7)$. Therefore, $3 y \leqslant 2 y+7$, and since $y>3$, thus $y=4,5,6,7$.
Upon verification, only $y=7$ satisfies $3 y \mid (2 y+7)$.
Substituting $x=3, y=7$ into the equation in the problem, we get $63=21 z, z=3$.
$\therefore$ The positive integer solutions that meet the conditions are $(x, y, z)=(1,3,5)$ or $(3,7,3)$.