Let p^2-p+1 =q^3

Small p

p=2, p^2-p+1 = 3,

p=3, p^2-p+1 = 7,

p=5, p^2-p+1 = 21, etc.

Since there are no solutions for p<7, 

Let (6n-1)^2-(6n-1)+1 = q^3

But (6n-1)^2-(6n-1)+1 = 3 (12n^2 - 6n + 1), never prime.

So let (6n+1)^2-(6n+1)+1 = 36n^2+6n+1  = q^3

The cubes mod9 are congruent to {0,1,8}, but the sequence 43,157,343,... is worth {7,4,1} mod9.

This adds a further constraint, requiring that 324n^2 + 18n + 1=q^3

True for {n,q} = {0,1}{1,7}

An investigation of 324n^2 + 18n + 1 mod 7,9,13,19 disclosed only 343, 63757,and 117307 as possible candidates up to 273007, the last number checked. But 63757 = 103*619, while 117307 is prime.

However, the similar Ljunggren equation (y^3-1)=x^3(y-1) can be rewritten (y-1) (y^2+y+1) = x^3 (y-1). Dividing through by (y-1) gives (y^2 + y + 1) = x^3 (identical to the given problem since (m-1)^2+(m-1)+1 = m^2-m+1) while the known sole solutions to (y^3-1)=x^3(y-1) are (18^3-1)=7^3(18-1) and ((-19)^3-1) = 7^3(-19)-1). As the greater includes the less, the same must be true of the given problem.

Accordingly, the only non-negative solutions to p^2-p+1=q^3 are {p,q} = {0,1}, {1,1}, and {19,7} and this is so irrespective of the primality of p or q.