$p^2-p+1=m^3$, then $p(p-1)=m^3-1=(m-1)(m^2+m+1)$ So either $p=3$ or $p$ divides one of $m-1$, $m^2+m+1$. $p=3$ obviously doesn't work. $(i)$ Let $p$ divide $m-1$, then $m^2+m+1$ divides $p-1$. so; $$p \leq m-1 \leq m^2+m+1 \leq p-1< p$$contradiction. $(ii)$ $p$ divides $m^2+m+1$ and $m-1$ divides $p-1$ This means that for some positive integer $k$, $m^2+m+1=pk$ and $p-1=(m-1)k$, so $$m^2+m+1=(mk-k+1)k=mk^2-k^2+k$$This is a quadratic equation for $m$. Then,$$\Delta = k^4-6k^2+4k-3$$must be a perfect square, but for $k>3$, $$(k^2-3)^2 \leq \Delta \leq (k^2-2)^2$$so it cannot be a perfect square. Plugging $k=1,2,3$ we get the solution $(p,m)=(19,7) $

parmenides51 wrote: Find all primes $p$ for which $p^2 - p + 1$ is a perfect cube. It's an old and classical problem from Balkan MO 2005 and has been posted many times before on this forum. See e.g. here. Please use the search function before posting!

As my interest is posting and adding more post collections in aops, such problem sets shall be posted inside Old High School Olympiads forum (the non geometry ones) to minimize number of reposts in HSO Forum. Those problem sets shall be available (as always even after being moved inside Aops Contest Collections) both in Contests to Add post and Old HSO index. I have the time to post problem sets missing from Aops Contest Collections, but not the time to search every one of those for possible reposts, so this shall be my workaround.