Example 14. Find the positive integer solutions of the equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.
Solution: Clearly, $x, y, z$ are all greater than 1, and at least one of $x, y, z$ is less than 4. Otherwise, if $x \geqslant 4, y \geqslant 4, z \geqslant 4$, then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{3}{4}$, which clearly contradicts the problem. Now let $x \leqslant y \leqslant z$, then $1<x<4$, so $x=2$ or 3. We perform the following trials:
(1) If $x=2$, then $\frac{1}{y}+\frac{1}{z}=1-\frac{1}{2}=\frac{1}{2}$,

i.e.,
$$
\begin{array}{c}
(y-2)(z-2)=4 . \\
\because z \geqslant y \geqslant x=2, \\
\therefore \quad\left\{\begin{array}{l}
y-2=1,2, \\
z-2=4,2 .
\end{array}\right.
\end{array}
$$
(2) If $x=3$, then $\frac{1}{y}+\frac{1}{z}=1-\frac{1}{3}=\frac{2}{3}$,

i.e., $(2 y-3)(2 z-3)=9$.
Similarly, we get $\left\{\begin{array}{l}y=3, \\ z=3\end{array}\right.$
Therefore, the solutions satisfying $x \leqslant y \leqslant z$ are
$$
\left\{\begin{array}{l}
x=2,2,3 \\
y=3,4,3 \\
z=6,4,3
\end{array}\right.
$$