<s>[INST] How to solve $p^2 - p + 1 = q^3$ over primes? 
How to solve $p^2 - p + 1 = q^3$ over primes? 

I checked that $p$ and $q$ must be odd and I don't know what to do next. I don't have any experience with such equations.
 [/INST] COMMENT.-Primes $p$ of the form $6x-1$ cannot be solution: in fact
$$(6x-1)^2-(6x-1)+1=36x^2-18x+3=3(12x^2-6x+1)$$ so $q=3$ which is not compatible with $12x^2-6x+1=3^2$.
It follows $p$ must be of the form $6x+1$ so we have
$$(6x+1)^2-(6x+1)+1=36x^2+6x+1=q^3$$ Solving the equation, where $X=6x$,
$$X^2+X+1-q^3=0$$ we have $$2X=-1+\sqrt{4q^3-3}$$ and the equation $$4q^3-3=z^2$$ seems to have $(z,q)=(37,7)$ as only solution which gives $\boxed{(p,q)=(19,7)}$.
Is it $(p,q)=(19,7)$ the only solution?
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