Diophantine equation $7b^2+7b+7=a^4$. 
Solve in positive integers $(a,b)$ the equation 
  $$7b^2+7b+7=a^4.$$

As the left side is divisible by $7$ we must have $7|a^4\implies 7|a$. Let $a=7c$, then the given equation reduces to $$b^2+b+1=343c^4.$$
I'm not sure how to proceed. Any hints or solutions are welcome.

If 
$$
b^2+b+1=343c^4
$$
then $b^2+b+1-343c^4=0$
$$
b=\frac{-1-\sqrt{1-4(1-343c^4)}}{2}=\frac{-1-\sqrt{1372c^4-3}}{2}
$$
So there is an integral solution iff there exist $d$ such as $1372c^4-3=d^2$, that is
$$
1372c^4-d^2=3
$$
A solution for the last equation is $c=1$, $d=37$. Hence $b=18$ or $b=-19$.
