here is my solution; i won't consider trivial cases: we assume that $ x > y \ge z$ note that we have $ x^2 | y^3 + z^3$ also we have $ \frac {x}{y^2z^2} + \frac {y}{z^2x^2} + \frac {z}{x^2y^2} = n$ by using inequality... $ \frac {x}{y^2z^2} + \frac {2}{x} > \frac {x}{y^2z^2} + \frac {y}{x^2} + \frac {z}{x^2} \ge n \implies \frac {x}{y^2z^2} > n - \frac {2}{x} \ge n - 1 \implies x > (n - 1)y^2z^2$ so we get $ x^2 > (n - 1)^2y^4z^4$. now the rest is easy by knowing that $ x^2 | y^3 + z^3 \implies (n - 1)^2y^4z^4 < x^2 \le y^3 + z^3$.

Without loss of generality, let $ z$ be the greatest of the variables. Then $ nx^2y^2z^2 = x^3 + y^3 + z^3 \leq 3z^3$, so $ nx^2y^2 \leq 3z$. Since $ x^2y^2 \leq nx^2y^2$, we have $ x^2y^2 \leq 3z$. On the other hand, $ z^2 | x^3 + y^3$, so $ x^3 + y^3 \geq z^2 \geq \frac {x^4}{y^4}{9} \geq \frac {x^3}{y^3}{9}$. Thus, $ 9x^3 + 9y^3 \geq x^3y^3$, so $ 81 \geq (x^3 - 9)(y^3 - 9)$. It is therefore to sufficient to check all $ (x, y)$ in $ (1, 2, 3, 4) \times (1, 2, 3, 4)$; the only $ x$ and $ y$ that yield solutions are $ (1,1)$, $ (1,2)$, and $ (2,1)$. Hence, our solution set with $ x \leq y \leq z$ is $ (1, 1, 1, 3), (1, 2, 3, 1)$.