Find all $n$ which $7(n^2 + n + 1)$ is perfect $4^{th}$ power. 

Find all positive integer $n$ , which $7(n^2 + n + 1)$ is perfect $4^{th}$ power.


What I tried
Let $7(n^2 + n + 1) = a^4$ $\to$ $ 7 | a$ and $a$ is odd.
We then get $(n^2 + n + 1) = 343k^4$ ; $k \in \mathbb Z$
Hence, $ 343 | n^3 - 1$. I’m stuck here
Please help! Thanks in advance.
Ps : This problem is from my teacher , in the topic of polynomial and it’s application.
 A: A partial solution. We may start by writing down the solutions of
$$ 7(n^2+n+1)=q^2 \longleftrightarrow 7(2n+1)^2-(2q)^2=-21$$
which are related (via $2n+1=m, q=7r$) to the solutions of the Pell equation
$$ m^2-7(2r)^2=-3$$
whose first solutions are given by $(m,r)\in\left\{(5,1),(37,7)\right\}$. The Pell equation $x^2-7y^2=-1$ has no solutions since $-1$ is not a quadratic residue $\!\!\pmod{7}$ and $x^2-7y^2=1$ has the fundamental solution $(x,y)=(8,3)$. It follows that the solutions of
$$ a^2-7b^2=-3 $$
are given by $(a,b)\in\{(2,1),(5,2),(37,14),(82,31),\ldots\}$ where the values of $a$ are given by OEIS A202637 and the values of $b$ are given by OEIS A202638. It follows that the values of $q$ are given by seven times the elements of the sequence OEIS A296377
$$ q_1=1, 7, 247, 1777, 62737, 451351, \ldots $$
fulfilling $q_n = 254 q_{n-2} - q_{n-4}$ for $n>4$. The last sequence $\!\!\pmod{7}$ has a period with length $14$ and the terms of the form $7k$ are the ones given by $q_{14s+2}$ and $q_{14s+13}$. It follows that
$$ 7(n^2+n+1)=q^2 $$
has an infinite number of non-trivial solutions, one of them being
$$ n=690845140450082,\quad q=1827804436088407 $$
and we still have to rule out the chance that (with the only exception of $q_2$) $q_{14s+2}$ or $q_{14s+13}$ is seven times a square.
