Example 12 The number of positive integer solutions to the system of equations $\left\{\begin{array}{l}x y+y z=63, \\ x z+y z=23\end{array}\right.$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
From the second equation, we get $z(x+y)=23$, and since 23 is a prime number, we can obtain $\left\{\begin{array}{l}z=1, \\ x+y=23\end{array}\right.$ or $\left\{\begin{array}{l}z=23, \\ x+y=1 \text {. }\end{array}\right.$ According to the problem, $x+y=1$ is not valid, so we discard it. Substituting $z=1$ into the given equations, we get
$$
\left\{\begin{array}{l}
x y+y=63, \\
x+y=23 .
\end{array}\right.
$$

Solving this, we get $(x, y)=(20,3),(2,21)$.
Therefore, the solutions to the original system of equations are
$$
(x, y, z)=(20,3,1),(2,21,1) .
$$