XVIII OBM

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.