Has the diophantine equation $z^2+zx=2x^2y$ infinitely many solutions over integers $x,y,z\geq 2$?

Question

After I did a reasoning involving even perfect numbers (see it if you want below my question) I am interested to know next

Question. Has $$z^2+zx=2x^2y\tag{D}$$ infinitelty many solutions $(x,y,z)$ over integers $x\geq 2$, $y\geq 2$ and $z\geq 2$? Many thanks.

To solve the question I did few reasonings like to try factorizations $z(z+x)$, or $x(2xy-z)$, and the code solve z^2+zx=2x^2 y, over integers using the Wolfram Alpha online calculator.

I don't know if our problem (D) was in the literature. I would like to know a parametric solution having infinitely many solutions, if you know it from the literature refer the literature, and I try to find it. If solve the Question is very hard, feel free to add your reasonings with this purpose to study this problem, or a remarkable heuristic if your experiments with a computer can provide us it.

---

I hope that these calculations are rights.

Claim. If there are infinitely many even perfect numbers, then (D) has infinitely many solutions*.

*As those that were stated, I say integer tuples $(x,y,z)$ with $x,y,z\geq 2$.

Proof. Let $M_p=2^p-1$ a Mersenne prime. The celebrated Lucas-Lehmer test, see [1] (is a free access journal) implies that there exist an integer $x\geq 2$ such that $$S_{p-2}=x\cdot M_p.\tag{1}$$ Thus denoting $z=S_{p-2}$ (see in [1] the definition, that is the sequence in the Lucas-Lehmer test), and since we know that an even perfect number can be written as $$y:=y_p=\frac{(M_p+1)}{2}M_p,\tag{2}$$ one gets solving $(2)$ for $M_p$ and doing the substitution in $(1)$, this
$$z=x\left(\frac{-1+\sqrt{1+8y}}{2}\right).\tag{3}$$ Write $(3)$ as $2z+x=x\sqrt{1+8y}$, and taking the square as $4z^2+4zx=8x^2y$.$\square$

Example. The tuple $(x,y,z)=(2,28,14)$ solves (D) since $$14^2+14\cdot 2-2\cdot 2^2\cdot 28=0,$$ and $x,y$ and $z$ are integers greater or equal than $2$.

References:

[1] Berrizbeitia and Luca and Melham, On a Compositeness Test for $(2^p+1)/3$, Journal of Integer Sequences, Vol. 13 (2010), Article 10.1.7.

Answer 1

One family of infinite solutions of the diophantine equation $z^2+zx=2x^2y$ is $$(x,y,z)=\left(n,\frac{mn(mn-1)}{2},n(mn-1)\right) $$ with $n,m\in\mathbb{Z}$. If $n,m\geq 2$, then $x,y,z\geq 2$. Your solution $(4,28,14)$ can be obtained by letting $n=2$ and $m=4$.

Answer 2

Take any $x \ge 2$ and let $z=2nx^2$ for some integer $n$. Then

$z^2 + zx = 2nx^2(2nx^2+x)$

so if you let $y=n(2nx^2+x)$ then $z^2 + zx = 2x^2y$.