Example 5 When $x \leqslant y \leqslant z$, find the positive integer solutions of the equation
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{7}{8}
$$

(2007, Taiyuan Junior High School Mathematics Competition)
Given $x>0$, and $x \leqslant y \leqslant z$, we know $\frac{1}{x} \geqslant \frac{1}{y} \geqslant \frac{1}{z}>0$.
Therefore, $\frac{1}{x}<\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{3}{x}$.
Thus, $\frac{1}{x}<\frac{7}{8} \leqslant \frac{3}{x}$.
Solving this, we get $\frac{8}{7}<x \leqslant \frac{24}{7}$.
Therefore, $x=2$ or 3.
When $x=2$, $\frac{1}{y}+\frac{1}{z}=\frac{3}{8}$.
Hence, $\frac{1}{y}<\frac{1}{y}+\frac{1}{z} \leqslant \frac{2}{y}$.
Thus, $\frac{1}{y}<\frac{3}{8} \leqslant \frac{2}{y}$.
Solving this, we get $\frac{8}{3}<y \leqslant \frac{16}{3}$.
So, $y=3,4,5$.
From $\frac{1}{y}+\frac{1}{z}=\frac{3}{8}$, we solve to get $z=24,8, \frac{40}{7}$ (discard).
When $x=3$, $\frac{1}{y}+\frac{1}{z}=\frac{13}{24}$.
Hence, $\frac{1}{y}<\frac{1}{y}+\frac{1}{z} \leqslant \frac{2}{y}$.
Thus, $\frac{1}{y}<\frac{13}{24} \leqslant \frac{2}{y}$.
Solving this, we get $\frac{24}{13}<y \leqslant \frac{48}{13}$.
So, $y=2$ (discard), 3.
From $\frac{1}{y}+\frac{1}{z}=\frac{13}{24}$, we solve to get $z=\frac{24}{5}$ (discard).
Therefore, the positive integer solutions to the original equation are $(x, y, z)=(2,3,24)$ and $(2,4,8)$.