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.

An elementary solution [albeit not using Pellian methods] is reasonably simple to obtain.
As you show, there is no solution when $y$ is even. Now assume $y$ is odd. Observe that $7 \mid (x^2+x+1)$ implies $x \equiv 2\!\pmod{7}$ or $x \equiv 4\!\pmod{7}$. 
In the first case, we substitute $x=7z+2$ into the equation and simplify to obtain
$$z(7z+5)=\bigl(7^\frac{y-1}{2}-1\bigr)\bigl(7^\frac{y-1}{2}+1\bigr).$$
Now our assumption of a positive solution with $y \ge 3$ implies there exist positive integers $a,b,c,d$ such that
\begin{align}
z &= ab, & 7^\frac{y-1}{2}-1 &= ac,  \\
7z+5 &= cd, & 7^\frac{y-1}{2}+1 &= bd.
\end{align}
The conclusion follows fairly directly via algebraic manipulations.
The proof in the second case is nearly identical.

EDIT:

Here's a hint to get most of the way through the proof… Writing $k=(y-1)/2$, we have
$$
\frac{z}{7^k+1} = \frac{a}{d} = \frac{7^k-1}{7z+5}.
$$
Solving each relation for $7^k$ gives
$$
7^k = \frac{dz-a}{a} = \frac{7az+5a+d}{d},
$$
and solving the second relation for $z$ yields
$$
z = \frac{a(2d+5a)}{d^2-7a^2}.
$$
Since $z$ is an integer by hypothesis, there must be a prime $p$ dividing $\gcd(d,a)$, or $(d^2-7a^2) \mid (2d+5a)$, from which we can quickly deduce $p \mid 3a$.
