If it is given that there are $3$ quantities, $x,y,z$ linked by the equality: $y^2 = x^3 +xz^4$ leading to the implications:
(i) y^2\mid x^3,
(ii) y^2\mid xz^4.
It is stated in the proof that: $p\mid y$ with $p=\gcd(x,z)$ i.e., $\exists X,Z\in \mathbb{Z},\,\,\, x=pX,z=pZ.$
Have two questions based on that :
$A$. How can it be implied that $p^2\mid y$? This is given in the proof as in the link to my earlier post, here.
To repeat, my question is why $p^3\mid y^2 \implies p^2\mid y$, if the title equality is assumed true.
$B$. Can I 'further' imply that $y\mid x$? $\,\,$If so, how?
Ok, I think I get it now: you have $y^2=x(x^2+z^4)$. What you can say is $x|y^2$ and $(x^2+z^4)|y^2$. You can't conlude that $y|x$ since that would mean $|y|\leq |x|$ but we already know that $|x|(x^2+z^4)=|y|^2\leq |x^2|$ which implies that $x^2+z^4\leq |x|$ since in your linked question you've taken $x,y,z \neq 0$. But this is trivially false as $x^2\geq |x|$ and $z^4\geq 1$ (since $z\neq0$).
If $p=\gcd(x,z)$, then $p^3|x(x^2+z^4)=y^2$. However, this doesn't imply $p^2|y$ since you have the counterexample $4^3|8^2$.