Hellenic Mathematical Olympiad

We write the given relation in the form $$x^2+y^2+xy=(6xy-x-y-z)z^2.$$

We observe that substituting $z$ by $6xy-x-y-z$, leaves invariant the right part of the equality. Thus, if $(x,y,z)$ is a solution of the given equation with $x>y>z$, then $(6xy-x-y-z,x,y)$, with $6xy-x-y-z>x>y$, is a solution, as well, because $$(6xy-x-y-z)-x=\frac{(x^2+y^2+xy)}{z}-x=\frac{(x(x-z)+y^2+xy)}{z}>0.$$

Since $(x,y,z)=(1,1,1)$ is a solution of the given equation, we obtain a new solution $(x,y,z)=(3,1,1)$, and then we find the solution $(x,y,z)=(13,3,1)$. Thus we conclude that the given equation has an infinite number of solutions.