A typical application of the infinite descent. We will show that $(0, 0, 0)$ is the only solution. Let us assume that $(x, y, z)$ is a solution not of that form. Then we notice that $x$ must be even, so we may let $x = 2k,$ for some $k \in \mathbb{Z}.$ Hence, our equation rewrites \[(2k)^{3}+2y^{3}= 4z^{3},\] or \[(-y)^{3}+2z^{3}= 4k^{3},\] so that $(-y, z, \frac{x}{2})$ is also a solution. Continuing like this, we get that $(-z, \frac{x}{2},-\frac{y}{2}),$ and so on and so forth, and we'll eventually get a solution containing an odd number, which is the desired contradiction.

for none of $ x,y,z=0$ we may assume that $ gcd(x,y,z) = 1$ and from $ a^3 = 0, - 1,1 (mod 9)$ will lead to the solution such that there is no $ (x,y,z)$ possible if none of $ x,y,z$ is equal to 0 if one of them is 0 the rest is trivial