The solution is $(x, y, z) = (2, 251, 252)$.
By the given, $x$ divides $2 \cdot 2012 = 2^3 \cdot 503$. Moreover, $x$ cannot be a multiple of $503$, by considering the exponent of $503$ in both terms of the equation. Similarly, by considering the powers of $2$ on both sides of the equation, we conclude that $x$ cannot be a multiple of $4$, thus $x = 1$ or $x = 2$. Thus, we have either
$$y^3 + z^3 = 2012(yz + 2) \text{ or } y^3 + z^3 = 503(yz + 1).$$
In either case, $503$ divides $y^3 + z^3$, so $y^3 \equiv -z^3 \pmod{503}$. Consequently, by Fermat's little theorem, we have $y^2 \equiv y^{504} \equiv y^{3 \cdot 168} \equiv z^{3 \cdot 168} \equiv z^2 \pmod{503}$. From $y^3 \equiv -z^3 \pmod{503}$ and $z^2 \equiv y^2 \pmod{503}$, we conclude that either both $y$ and $z$ are divisible by $503$ or $y \equiv -z \pmod{503}$. Hence in all cases $y+z$ is a multiple of $503$. Write $y+z=503k$ with $k$ a positive integer.
Next, we cannot have $y = z$, as it would then follow that either $y^2 + 1$ or $y^2 + 2$ divides $2y^3$, which forces $y \le 1$. For $(y-z)^2 = 1$, say $y = z + 1$ the first equation has no solution because $y+z$ must be even, while the second one has only the solution $(y, z) = (251, 252)$, as in this case the equation simply becomes $x + y = 503$. We now prove that we cannot have $(y-z)^2 \ge 4$. Indeed, if this happened, write $y+z = 503k$ and note that $y^2 - yz + z^2 > yz + 2$. Thus for the first equation we would get $k < 4$, while for the second one $k < 1$. Consider the first equation with $k < 4$. As $y+z$ is even, $k$ is even and so $k = 3$. But then $3$ divides both $y+z$ and $yz+2$, so $3$ divides $y^2+1$, which is impossible.
Thus, the only solution is $(2, 251, 252)$.