<s>[INST] Why does the diophantine equation $x^2+x+1=7^y$ have no integer solutions? This following Problem is from Pell equation chapters exercise

Let $y>3$ positive integer numbers, show that following diophantine equation
  $$x^2+x+1=7^y\tag{1}$$ has no integer solutions.

I tried write the equation
$$(2x+1)^2+3=4\cdot 7^y$$
if $y=2k$ then we have
$$(2\cdot 7^k+2x+1)(2\cdot 7^k-2x-1)=3$$
this case has no integer.
But $y$ is odd number, How to prove equation  (1)  has no  integer solutions for $x,y (y>3)$? Any help would be appreciated.
 [/INST] We work in the Eisenstein integers. The equation then becomes$$(x + 1 + \omega)(x - \omega) = 6^y.$$Suppose $d\, |\, (x + 1 + \omega)$ and $\text{ }d\,|\,(x - \omega)$.Then we have $d\,|\,(1 + 2\omega)$, so $N(d)\,|\,3$. However, we also have $N(d)\,|\,7^{2y}$, so $N(d) = 1$. On the other hand, $7 = (-2 - 3\omega)(1 + 3\omega)$ and $-2 - 3\omega$, $1 + 3\omega$ are relatively prime. Thus, it follows that $x + 1 + \omega$ and $x - \omega$ are $e(-2-3\omega)^y$ and $f(1 + 3\omega)^y$ in some order, where $e$ and $f$ are units $\pm1$, $\pm\omega$, $\pm1\pm\omega$ with $ef = 1$. We just need to show that if$$a + b\omega = f(1 + 3\omega)^y,$$then $|b|$ can never be $1$, which would solve the problem. Indeed, let$$A = 1 + 3^3\binom{y}{3} + \dots,\text{ }B = 3\binom{y}{1} + \dots,\text{ }C = 3^2\binom{y}{2} + \dots.$$We have$$a + b\omega = f(A - C + \omega(B - C)),$$so we want to show none of $|A - B|$, $|B- C|$, $|C - A|$ are one. However, this is a relatively easy task for $y > 3$, so we are done.
</s>