SAMC

Write the equation $p^2-p+1=x^3$ as $$p(p-1)=(x-1)\left(x^2+x+1\right).$$ Because $p>x$, $p$ divides $x^2+x+1$ so $x^2+x+1=kp$ and $k(x-1)=p-1$, for some positive integer $k$. It follows that $$k^2(x-1)+k=x^2+x+1$$ and consider the following cases:

Case 1. If $k^2\le x+1$, then $$k^2(x-1)+k\le x^2-1+k\le x^2-1+\sqrt{x+1}<x^2+x+1,$$ not possible.

Case 2. If $k^2\ge x+3$, then $$\begin{array}{rl}{k}^2(x-1)+k& \ge (x-1)(x+3)+k\ge (x-1)(x+3)+\sqrt{x+3}\\ & =x^2+2x-3+\sqrt{x+3}>x^2+x+1\end{array}$$ not possible.

Therefore $x+1<k^2<x+3$, hence $k^2=x+2$. This implies $$(x+2)(x-1)+k=x^2+x+1,$$ that is $k=3$. It follows $x=7$ and $p=19$.