Example 8 Try to find all positive integers $n$ such that the equation
$$
x^{3}+y^{3}+z^{3}=n x^{2} y^{2} z^{2}
$$

has positive integer solutions.
Solution: Without loss of generality, let $x \leqslant y \leqslant z$. Clearly, $z^{2} \mid\left(x^{3}+y^{3}\right)$, so, $z^{2} \leqslant x^{3}+y^{3}$.
But $x^{3} \leqslant x z^{2}, y^{3} \leqslant y z^{2}$, hence we have
$$
z=n x^{2} y^{2}-\frac{x^{3}+y^{3}}{z^{2}} \geqslant n x^{2} y^{2}-(x+y) \text {. }
$$

Thus, we have
$$
\begin{array}{l}
x^{3}+y^{3} \geqslant z^{2} \geqslant\left[n x^{2} y^{2}-(x+y)\right]^{2}, \\
n^{2} x^{4} y^{4}<2 n x^{2} y^{2}(x+y)+x^{3}+y^{3}, \\
n x y<2\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{n x^{3}}+\frac{1}{n y^{3}} .
\end{array}
$$

Notice that the left side of inequality (1) increases with $x, y$, while the right side decreases with $x, y$, i.e., the upper bound of $n x y$ does not exceed $2\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{n x^{3}}+\frac{1}{n y^{3}}$, and the lower bound of $2\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{n x^{3}}+\frac{1}{n y^{3}}$ is no less than $n x y$. It is easy to see that $x=1$.
Substituting $x=1$ into equation (1) yields
$n y<2+\frac{2}{y}+\frac{1}{n}+\frac{1}{n y^{3}}$.
Equation (2) can only hold when $y \leqslant 3$.
But $z^{2} \mid\left(1+y^{3}\right)$, so it can only be $y=1,2(y=3$ when, $z^{2} \mid 28, z^{2}=1,4$, which contradicts $z \geqslant y$), accordingly, $z$ $=1,3$. At this point, $n=3,1$.