We replace $2010$ by $p-1$, for some odd prime. Subtracting the first equation from the second, we obtain
$$(x-y)(y-z)(z-x) = -p(p-1)$$
We have $(x-y)+(y-z)+(z-x)=0$ and $(x-y)(y-z)(z-x)<0$, so precisely two of them are positive. Assume that $x-y>0$ and $y-z>0$. Without loss of generality, suppose $x-y \leq y-z$. Because $p$ is a prime, the only possibility is
$$x-y=1, \quad y-z=p-1, \quad z-x=-p$$
Then $x=y+1$, $z=y-(p-1)$, and the first equation reduces to
$$y\left[3y^{2} + 3(p-2)y + (p-2)^{2}\right]=0$$
The only solution is $y=0$, implying $x=1$ and $z=p-1$.
The solutions $(x, y, z)$ are $(1,0,1-p)$, $(1-p, 1,0)$ and $(0,1-p, 1)$. We have $p=2011$, hence the desired triples are $(1,0,-2010)$, $(-2010,1,0)$, and $(0,-2010,1)$.