Example 9 If positive integers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5} \text {, }
$$

find the maximum value of $x_{5}$.
Let's assume $x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}$. Then
$$
\begin{array}{l}
1=\frac{1}{x_{2} x_{3} x_{4} x_{5}}+\frac{1}{x_{1} x_{3} x_{4} x_{5}}+\frac{1}{x_{1} x_{2} x_{4} x_{5}}+ \\
\frac{1}{x_{1} x_{2} x_{3} x_{5}}+\frac{1}{x_{1} x_{2} x_{3} x_{4}} \\
\leqslant \frac{3}{x_{4} x_{5}}+\frac{1}{x_{5}}+\frac{1}{x_{4}} \\
\Rightarrow \frac{3+x_{4}+x_{5}}{x_{4} x_{5}} \geqslant 1 \\
\Rightarrow x_{4} x_{5}-x_{4}-x_{5} \leqslant 3 \\
\Rightarrow\left(x_{4}-1\right)\left(x_{5}-1\right) \leqslant 4 \text {. } \\
\end{array}
$$

If $x_{4}=1$, then $x_{1}=x_{2}=x_{3}=x_{4}=1$.
Thus, $4+x_{5}=x_{5}$, which is not valid.
If $x_{4}>1$, then $x_{5} \leqslant 5$, when $x_{5}=5$, we can get the array $(1,1,1,2,5)$ that satisfies the condition.
Therefore, the maximum value of $x_{5}$ is 5.