Example 6 Try to find all positive integers $n$ such that the equation $x^{3}+y^{3}+z^{3}=n x^{2} \cdot y^{2} \cdot z^{2}$ has positive integer solutions.
Let's assume $x \leqslant y \leqslant z$, it is clear that $z^{2} \mid x^{3}+y^{3}$, hence $z^{2} \leqslant x^{3}+y^{3}$, but $x^{3} \leqslant x z^{2}, y^{3} \leqslant y z^{2}$.
Transforming $x^{3}+y^{3}+z^{3}=n x^{2} y^{2} z^{2}$, we get
$$
z=n x^{2} y^{2}-\frac{x^{3}+y^{3}}{z^{2}} \geqslant n x^{2} y^{2}-(x+y) \text {, }
$$

Thus, $x^{3}+y^{3} \geqslant z^{2} \geqslant\left[n x^{2} y^{2}-(x+y)\right]^{2}, n^{2} x^{4} y^{4}<2 n x^{2} y^{2}(x+y)+x^{3}+y^{3}$,
$$
n x y<2\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{n x^{3}}+\frac{1}{n y^{3}} \text {. }
$$

It is easy to see that $x=1$, otherwise if $x \geqslant 2$, and $y \geqslant 2$, then the left side of (1) is no less than 4, while the right side is clearly less than 3, (1) will not hold.
Given $x=1$, we have $n y<2+\frac{2}{y}+\frac{1}{n}+\frac{1}{n y^{3}}$.
The above inequality can only hold when $y \leqslant 3$, but $z^{2} \mid\left(1+y^{3}\right)$. Therefore, it can only be $y=1,2$ (when $y=3$, $z^{2} \mid 28, z^{2}=1,4$ contradicts $z \geqslant y$), accordingly, $z=1,3$.
In summary, we can only have $x=y=1$ or $x=1, y=2, z=3$, thus $n$ can only be 3,1.