Triplets based equation

Question

Let $p \ge 7$ be a prime number. Find the triples $(x, y, z)$ in $\mathbb{Z}$ such as $xyz$ is not equal to zero, $\gcd (x, y, z) = 1$ and $x^p + 2y^p = z^2$. I want triplets and proof/generalization. The reason for asking here, I am in position to construct equations and finding solutions by trail method. I am not in position to construct a proof or good generalizations. I hope, with your help, I can end.

Answer 1

There is a compiled list titled "SOME OPEN PROBLEMS ABOUT DIOPHANTINE EQUATIONS" and this problem happens to be listed as Problem#16 on page 3. Check it here

(Originally thought it was Problem #15, but I stand corrected it is indeed Problem#16).

Answer 2

For $p=7$, let $a$ be a positive integer. Then
$$\begin{align*} (7a^2)^7 + 2(21.a^2)^7 &= 7^7a^{14} + 7^7a^{14}.2.3^7\\ &=7^7a^{14}(1+2.3^7)\\ &=7^7.a^{14}.4375\\ &=7^6a^{14}.175^2\\ &=(60025a^7)^2, \end{align*} $$ so there are an infinite number of triplets $(7a^2, 21a^2, 60025a^7).$

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On re-reading the question, I see the gcd($x$, $y$, $z$)=1 constraint which kind of knobbles my answer. Ho hum.