A^3=b^3 implies a=b only when p is 2 mod 3. Divide by b^-3 and you get z^3=1 mod p as 3 is prime then ordpz = 3 so 3 | p-1. Note if b is 0 mod p then so is a. The other case is ab^-1=1 so the inverse of b is as inverse i.e a=b which is a simple case.

The condition may be restated as follows: a prime $p$ is not good if and only if there exist integers $a, b$ with $p \mid a^2 + ab + b^2$. Note that $$a^2 + ab + b^2 \equiv 0 \pmod{p} \implies (a + b)^2 \equiv ab \pmod{p} \hspace{3mm} \text{and} \hspace{3mm} (a - b)^2 \equiv -3ab \pmod{p}$$whence $\left(\frac{-3ab}{p}\right) = \left(\frac{ab}{p}\right)\left(\frac{3}{p}\right)\left(\frac{-1}{p}\right)$. If $p \equiv 1 \pmod{4}$ then $\left(\frac{-1}{p}\right) = 1$ whence $$\left(\frac{p}{3}\right)\left(\frac{3}{p}\right) = (-1)^{\frac{p-1}{2}} = 1$$so $\left(\frac{p}{3}\right) = 1 \implies p \equiv 1 \pmod{3}.$ By Dirichlet's theorem, there are infinitely many not good primes satisfying $p \equiv 1 \pmod{3}$ and infinitely many good primes satisfying $p \equiv 2 \pmod{3}$, as desired.

Why is the condition that $(a+b)^2 \equiv ab (\mod p)$ trivially true from $-3$ being a quadratic residue of $p$ (I understand the reverse; i.e. the implication that $-3$ is a quadratic residue $(\mod p)$.