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 Diophantine equation with prime powers
Not a complete answer but an indication of what is known:
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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 $n =0, 1, \dots$. 
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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.)
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