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.
 A: As I found while researching
this
Mathoverflow question, there is a Magma function
IntegralQuarticPoints
for finding all integer points on a curve $y^2 = Q(x) = ax^4+bx^3+cx^2+dx+e$
given also a rational point.
Here we have $x=a$ and $Q(x) = 28x^4-147$ (the discriminant of
$7(b^2+b+1) = x^4$ as a quadratic in $b$), and we know the rational point
$(x,y) = (7,259)$; so we ask Magma
time IntegralQuarticPoints([28,0,0,0,-147],[7,259]);

and get output ending with
[
    [ -7, -259 ],
    [ 7, -259 ],
]
Time: 0.330

(the Time is reported in seconds).  So there's a proof that the
known solutions with $a=\pm 7$ are the only ones, but it might use
some rather sophisticated mathematics along the way.  Some of the
preceding output indicates that the same equation $y^2 = 28x^4-147$
with rational (rather than integral) variables is an elliptic curve with a
rational $2$-torsion point and rank $1$, which I think means that
there's probably a proof that's not as advanced as whatever
Magma is using but still not simple.  (When a Diophantine
equation has no solution at all, there is often an elementary proof;
but once there's a nontrivial solution it's usually hard to find an elementary 
proof that allows that solution (as it must) but excludes all others.)
A: 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 $b=-19$.
