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 $x^{2}+x+1 = k p$ 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 $k^{2} \leq x+1$, then
$$k^{2}(x-1) + k \leq x^{2} - 1 + k \leq x^{2} - 1 + \sqrt{x+1} < x^{2} + x + 1,$$
not possible.

Case 2. If $k^{2} \geq x+3$, then
$$\begin{aligned} k^{2}(x-1) + k & \geq (x-1)(x+3) + k \geq (x-1)(x+3) + \sqrt{x+3} \\ & = x^{2} + 2x - 3 + \sqrt{x+3} > x^{2} + x + 1 \end{aligned}$$
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 $p = 19$.