To bound or not to bound!

Suppose $p$ and $q$ are natural numbers such that $x=zp$ and $y=zq$ . It follows that $zp+z^2 q^2+z^3=z^3 pq \\ p+z q^2 +z^2=z^2 pq$ Therefore $z|p$ and there is a natural number $r$ such that $p=rz$ . Put this in the last equation: $rz+z q^2+z^2=z^3 rq \\ r+q^2+z=z^2 rq \\ q^2-q(z^2 r)+(r+z)=0$ In order for getting integer values for $q$ , discriminant of the last equation (a quadratic in terms of $q$ ) $\Delta=(z^2 r)^2 -4(r+z)$ must be a perfect square, but $\Delta<(z^2r)^2$ Hence $\Delta \le (z^2 r-1)^2 \\ (z^2 r)^2-4(r+z) \le (z^2r-1)^2 \\ 1+4(r+z) \ge 2z^2 r \\ 1+4z \ge r(2z^2-4) \ge 2z^2-4 \\ 2z^2-4z-5 \le 0$ As a result $z\le 2$ . So $z=1$ or $z=2$ . If $z=1$ then $\Delta=r^2-4r-4=(r-2)^2-8$ Since $\Delta$ is a perfect square, difference of two squares are $8$ , hence squares are $1$ and $9$ . So $r-2=3$ and $r=5$ . In this case, $5+q^2+1=5q \\ q^2-5q+6=0$ Thus $q=2$ or $q=3$ which leads to two solutions: $(5,2,1), (5,3,1)$ .

If $z=2$ then $\Delta=(4r)^2-4(r+2)$ must be a perfect square, but $4r^2-(r+2) \le (2r-1)^2=4r^2-4r+1 \\ 4r-1 \le r+2 \\ r \le 1$ Therefore $r=1$ . Now we can write $1+q^2+2=4q \\ q^2-4q+3=0$ In conclusion $q=1$ or $q=3$ which gives us two solutions: $(4,2,2), (4,6,2)$ .