Brazilian Mathematical Olympiad

We can regard $x^2 - 3yz \cdot x + (y^2 + z^2) = 0$ as a quadratic in $x$. So if $1 \le x \le y \le z$ is a solution, then so is $y, z, 3yz - x$.

Also we have $3y \ge 3$, so $3yz - x \ge 3yz - z \ge 2z > z$. So $1 \le y \le z < 3yz - x$, and the new solution has larger largest element.

So starting with the solution $x = y = z = 1$ and repeating, we get infinitely many solutions.