If $x>y\ge1$ are integers, with $x$ even and $y$ odd, and $x^4+4x^3y-4xy^3-4y^4$ is a square, must $y \mid x$?

Question

The title says it all: Let $x$ and $y$ be integers, with $x$ even and $y$ odd, such that $x > y \ge 1$ and $$x^4+4x^3y-4xy^3-4y^4 = w^2$$ for some integer $w$. Does this force $y \mid x$?

Brute force calculations seem to support that conjecture. In fact, they suggest the stronger conjecture $x=2y$.

I’ve poked around trying to prove something, but can’t figure out the magic incantation. Any advice/help would be appreciated.

Answer 1

It seems that the pair $x=951017531851446281396498, y=780923760568941116026369$ is a counterexample.

Answer 2

I also tried a brute force approach, and it seems that for $x=265$ and $y=222$, you do get a square $143736121=11989^2$, but $222 \not\mid 265$.

[Edit: in a previous version of the question, the conditions on the parity of $x$ and $y$ were not there.]