$x^2+x+1$ is the cube of a prime. Please help me find all natural numbers $x$ so that $x^2+x+1$ is the cube of a prime number.(Used in here)

This is never a complete solution, but larger than the size that a comment can afford
If $x^2+x+1=p^3$ where $p$ is prime.
Clearly, $p\ne2$
As $p$ divides $x^2+x+1,p$ will divide $(x-1)(x^2+x+1)=x^3-1$
$\implies x^3\equiv1\pmod p\implies ord_px=3\implies 3$ divides $p-1$
So, prime $p=3q+1$ for some natural number $q$
As, $p>2$ i.e., odd, $p$ can be written as $6r+1$ for some natural number $r$
Now, $x^2+x=p^3-1=(6r+1)^3-1=18r(12r^2+6r+1)$
Observe that $(18r,12r^2+6r+1)=1$
As $(x,x+1)=1,$
$x,x+1$ can be $18r,12r^2+6r+1$
$\implies 12r^2+6r+1-(18r)=x+1-x=1$
$\implies  12r^2-12r=0\implies r=1$ as $r>0$
Now, we can test for other obvious combinations unless we can factorize $12r^2+6r+1$
$x=2r,x+1=9(12r^2+6r+1)$
$x=9r,x+1=2(12r^2+6r+1)$
etc.
