Proof for $x=2k$ $y=\frac{-x^3+-\sqrt{x^6-4z^2-8}}{2}\implies x=2k \implies y=-4k^3+-\sqrt{16k^6-z^2-2}\implies 16k^6-z^2-2=m^2\implies 3z^2+2 \equiv 0,1(mod 4)$ $z^2\equiv 0,1 (mod 4)\implies z^2 \equiv 1 (mod 4)$ Case 1: $x=2k ,y=4a+3$ $2+8k^3y+y^2+(4a+3)=0 \implies y^2+11 \equiv 0 (mod 8)\implies y^2\equiv 5 (mod 8)$ no solution. Case2:$x=2k,z=4a+1$ $2+8k^3y+y^2+(4a+1)^2\implies y^2+3\equiv 0 (mod 8)$ no solution

crezk wrote: $y=\frac{-x^3+-\sqrt{x^6-4z^2-8}}{2}\iff x=2k $ Nope. In fact it is easy to see that $x$ must be odd (and then I don't think that you can get an easy contradiction modulo $8$).

Thank you!I could have been more careful

let $x=2k+1\implies (2k+1)^6-4z^2-8\equiv-4z^2-7\equiv 2z^2-7\equiv 1,3 (mod 6)\iff 2z^2-7\equiv 1 (mod 6) \vee 2z^2-7\equiv 3 (mod 6)$ $\implies z^2\equiv 4 (mod 6) \vee z^2\equiv 5 (mod 6)$ also second impossible also $y+y^2+z^2 \equiv 0 (mod 2)\implies z^2\equiv 0 (mod 2)$ then $z$ is even $\implies z=6m+2,z=6m+4$ maybe this substitution works for main equation

Pure modular reasoning cannot possibly work for this problem. Indeed, putting $y=1$, note that we in particular need to prove that $z^2+3=x^3$ has no integer solutions. This is an instance of Mordell's equation, which indeed does not have integer solutions, but it is easy to show that it has solutions modulo all integers $n$, so the argument definitely has to use another ingredient!

Here is my attempt at a solution : We can view the equation as a quadratic in $y$ and set the discriminant equal to $k^2$. $$k^2=\Delta =x^6-4(z^2+2)$$This simplifies to $x^6=k^2+4z^2+8$. We first have that $v_2(k^2+4z^2+8)\leq v_2(4k'^2+4z^2+8)=$ $2+v_2(k'^2+z^2+2)\leq 4$ by evaluating modulo $8$. Thus we have that $x$ must be odd. Re-write the equation as $$(x^2-2)(x^4+2x^2+8)=k^2+(2z)^2$$Notice that if $p| (x^2-2)$ then $p=\pm 1\pmod{8}$. There must exist a prime $p\equiv -1\pmod{8}$ dividing $x^2-2$ as $x^2-2$ is not equivalent to $1\pmod{8}$. As $(x^4+2x^2+8)\equiv 16 \pmod{x^2-2}$ we must have that $v_p(x^6-8)= 1$. But this implies that $x^6-8$ can not be written as the sum of two squares.

Nice! The second factor in your factorization should be $x^4+2x^2+4$, I think, but fortunately, the argument still works: You get $x^4+2x^2+4 \equiv 12 \pmod{x^2-2}$, but $p=3$ cannot occur. Also, I think it would suffice to consider just primes $p \equiv 3 \pmod{4}$ (instead of mod $8$). So a slightly cleaner version of your argument would be: Note that $x^2-2$ and $x^4+2x^2+4$ are coprime (as their gcd would divide $12$, but $x^2-2$ is coprime to $2$ and to $3$), hence both would need to have all prime divisors $p \equiv 3 \pmod{4}$ an even number of times, but this contradicts $x^2-2 \equiv 3 \pmod{4}$.