A Diophantine equation with prime powers Let $p$ and $q$ be prime numbers such that $p^2+p+1=3q^a$: is it true that $a=1$?
This specific equation appears when computing order components of finite groups.
 A: Another counterexample for $a=2$: $p=2288805793$, $q=1321442641$.
There are only 2 counterexamples in primes below $10^{5000}$.
A: Not a complete answer but an indication of what is known:
The solutions to your diophantine equation known due to Nagell. I have in front of me Paulo Ribenboim's "My Numbers, My Friends" from which I quote:  

Theorem. If $m > 2$, the only non-zero solutions of $X^2 + X + 1 = 3 Y^m$ are $x = 1$ and $x = -2$. If $m = 2$, there are also the solutions
  $$x = \pm \frac{\sqrt{3}}{4}\left((2+\sqrt{3})^{2n+1} - (2-\sqrt{3})^{2n+1}\right) - \frac{1}{2}$$
  for $n =0, 1, \dots$. 

The $m = 2$ case is quite clear: multiply through by $4$ and note that the given equation is equivalent to 
$$(2X+1)^2 - 3(2Y)^2 = -3$$
Put $X' = 2Y$ and $Y' = (2X+1)/3$ so that you get $$X'^2 - 3Y'^2 = 1$$ and $2 - \sqrt{3}$ is the fundamental unit in the real quadratic field $\mathbf{Q}(\sqrt{3})$. 
The reference from Paulo's book is: 

T. Nagell. Des équations indéterminées $x^2 + x + 1 = y^n$ et $x^2 + x + 1 = 3y^n$. Norsk Mat. Forenings Skrifter, Ser. I, 1921, No. 2, 14 pages. (= 1921a at the end of Chapter 7.)

A: $$313^2 + 313 + 1 = 3\times181^2$$
