<s>[INST] Where is the mistake in this incorrect proof in Eisenstein integers? The Diophantine equation is $n^2 + n + 1 = m^3$ my attempt to solve it shows there is no solutions to this equation, but in fact there are four. I could not find my mistake so I hope someone could point it out please.

Factoring in the Eisenstein integers $\mathbb{Z}[\omega]$ with $\omega^2 + \omega + 1 = 0$: $n^2 + n + 1 = (n - \omega)(n + 1 + \omega)$.
If $d$ is a common divisor then $d$ also divides their difference $1 + 2 \omega$ which is a prime.
If the greatest common divisor is $1$ then both numbers are cubes, but that is impossible because $(a + b \omega)^3 = (a^3 - 3 a b^2 + b^3) + 3 (a^2 b - a b^2)\omega$ and the $\omega$ term of the factors are not multiples of $3$.
If the greatest common divisor is $1 + 2 \omega$ then $(n - \omega)(n + 1 + \omega) = -3 \frac{n - \omega}{1+2 \omega}\frac{n + 1 + \omega}{1 + 2 \omega}$ so either $n - \omega$ or $n + 1 + \omega$ is $1 + 2 \omega$ times a cube, but that would imply $2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 = (a-2b)(a+b)(2a-b) = \pm 1$ which is impossible.

Edit: Does he make the same mistake here Link ?
 [/INST] From 
$$\mbox{cube}= -3 \frac{n - \omega}{1+2 \omega}\frac{n + 1 + \omega}{1 + 2 \omega}$$ doesn't the inference 
"so either $n−\omega\ $ or $n+1+\omega\ $ is $1+2\omega\ $ times a cube"
also need $-3$ to be prime? This is not true in the Eisenstein integers.
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