$n$ can be $1$ or $3$.

WLOG assume $x \ge y \ge z$. Then we have
$$3x^3 \ge x^3 + y^3 + z^3 = nx^2y^2z^2,$$
which gives $x \ge \frac{n}{3}y^2z^2$.

Next, we rewrite the equation as $y^3 + z^3 = x^2(ny^2z^2 - x)$. Since the left-hand side is positive, we have $ny^2z^2 - x \ge 1$. Thus,
$$2y^3 \ge y^3 + z^3 \ge \left(\frac{n}{3}y^2z^2\right)^2 (1) = \frac{n^2}{9}y^4z^4.$$
This yields $18 \ge n^2yz^4$. There is no positive integer solution when $n \ge 5$. When $n = 4$, we must have $y = z = 1$. The given equation becomes $x^3 + 2 = 4x^2$. This shows $x^2 \mid 2$, and hence $x = 1$. But this is not a solution.

For $n = 3$, we can take $x = y = z = 1$.
For $n = 2$, we have $18 \ge 4yz^4$. So $z = 1$ and $y \le 4$. Note that $x^2 \mid y^3 + z^3 = y^3 + 1$.
As $x \ge y$, $(x, y)$ can only be $(1, 1)$, $(3, 2)$. It is routine to check that none of these is a solution to $x^3 + y^3 + 1 = 2x^2y^2$.
For $n = 1$, we can take $(x, y, z) = (3, 2, 1)$.
Thus, $n$ can only be $1$ or $3$.