Show that the equation $y^2 = x^3 +xz^4$ has no solutions in nonzero integers $x,y,z$. 
Show that the equation $y^2 = x^3 +xz^4$ has no solutions in nonzero integers $x,y,z$. 

There is given hint: Suppose that there is a solution. First show that it can be reduced to a solution satisfying $\gcd(x,z)=1$. Then use the fact that $x^3+ xz^4= x(x^2+z^4)$ is a perfect square to show that there are no solutions other than $x=y=0$.
The hint tells to prove by contradiction approach. So, $y\mid x,z\implies y\mid p$, where $p=(x,z)$. Hence, $\exists X,Z\in \mathbb{Z},\,x=Xp, z=Zp\implies y^2= X^3p^3 +XpZ^4p^4$$\implies y^2= Xp^3(X^2+Z^4p^2)$.

Is it valid to conclude from the last line, any of the below two separate conclusions : 
1. The set of common divisors of $y^2, x^3, xz^4$ are the same; hence let $gcd(y^2, x^3, xz^4)=p'$
2.$\,\,p^3\mid y^2\implies \exists y=p^2Y, \,\,\,\,p^2\mid y$
 A: This is not much more than a lengthy remark, but it's worth posting as an answer, I think, because it shows that proving this is at least as hard as proving the strong version of Fermat's Last Theorem for $n=4$. That is, suppose that $a^4+b^4=c^2$ with $abc\not=0$. Then, letting $x=a^2$, $z=b$, and $y=ac$, we first have $x^2+z^4=c^2$, and thus $x(x^2+z^4)=xc^2=(ac)^2=y^2$ with $xyz=a^3cb\not=0$. So if $y^2=x^3+xz^4$ has no solutions with $xyz\not=0$, then $a^4+b^4=c^2$ has no solutions with $abc\not=0$ either.
A: For starers, we must find $x,y,z$ all positive integers. 
There are two possibilities : either $x$ is a square, say $x=A^2$ or otherwise $x=a^2b$ in which b is a square free integer.
Case 1. $y^2=A^2(A^4+z^4) \implies (y/A)^2 = Y^2 = A^4+z^4$ which is impossible due to the Fermat's LT (strong version for $n=4$).
Case 2. $y^2=a^2b(a^4b^2+z^4) \implies (y^2/a^2) = b(a^4b^2+z^4) \implies b|(a^4b^2+z^4) \implies b|z^4 \implies b|z$. So, $(y/ab)^2 = b(a^4+z_1^4b^2)$ in which $z=z_1b$. Again, $b|a^4$ which implies $b|a$. Thus, $(y/ab^2)^2 = (a_1^4b^3+z_1^4b) = b(a_1^4b^2+z_1^4)$ in which $a=a_1b$. So every time we find a new positive integer satisfying in the non-ending sequence $y>y_1=y/a>y_2=y/(ab)> \dots$ unless $a=1=b$ thus  $x=1$. But if $x=1$ $y^2=1+z^4$ has positive integer solutions contradicting the mentioned Fermat's Theorem.     
