= Solution
Factor the equation in the Eisenstein integers:
$$
x^2-x+1=(x+\omega)(x+\omega^2)=y^3.
$$
Let $\pi=1-\omega$. Its norm is $3$, and
$$
\pi^2=-3\omega,
$$
so $3$ is a unit times $\pi^2$. Suppose $x\equiv2\pmod3$, say $x=3k+2$. Since $\omega=1-\pi$,
$$
x+\omega=3(k+1)-\pi.
$$
The first term is divisible by $\pi^2$ and the second has <discrete valuation> exactly one, so $v_\pi(x+\omega)=1$. Conjugation gives $v_\pi(x+\omega^2)=1$ as well. Consequently
$$
v_\pi(x^2-x+1)=2.
$$
But $v_\pi(y^3)=3v_\pi(y)$ is divisible by three, a contradiction. Hence
$$
\boxed{x\not\equiv2\pmod3}.
$$
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