Find all positive integers $k$ such that the equation
$$
x^{2}+y^{2}+z^{2}=k x y z
$$

has positive integer solutions $(x, y, z)$.
Let $(a, b, c)$ be a solution, and among all solutions, $(a, b, c)$ minimizes $x+y+z$. Assume $a \geqslant b \geqslant c$. By Vieta's formulas, we have $d=\frac{b^{2}+c^{2}}{a}=k b c-a$, which means $x^{2}+b^{2}+c^{2}=$ $k b c x$ (2) has positive integer solutions. Moreover, $(b, c, d)$ is a solution to (1). Thus, by the minimality of $(a, b, c)$,
$$
\begin{array}{l}
a \leqslant d, a^{2} \leqslant a d=b^{2}+c^{2}<(b+c)^{2}, a<b+c, \\
k=\frac{a^{2}+b^{2}+c^{2}}{a b c}=\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}<\frac{b+c}{b c}+2=\frac{1}{b}+\frac{1}{c}+2 \leqslant 4,
\end{array}
$$

Therefore, $k=1, 2, 3$.
However, when $k=2$, $x^{2}+y^{2}+z^{2}$ is even, so at least one of $x, y, z$ is even, and $k x y z$ is divisible by 4, $x^{2}+y^{2}+z^{2} \equiv 0(\bmod 4)$, which implies $x, y, z$ are all even. Let $x=2 x_{1}, y=2 y_{1}$, $z=2 z_{1}$. Substituting into (1) gives $x_{1}^{2}+y_{1}^{2}+z_{1}^{2}=4 x_{1} y_{1} z_{1}$. This has no solutions.

Thus, $k=1$ or $3$. $x=y=z=1$ is a solution to the equation $x^{2}+y^{2}+z^{2}=3 x y z$. $x=y=z=3$ is a solution to the equation $x^{2}+y^{2}+z^{2}=x y z$.