$p|n^3-1 \implies p|(n-1)(n^2+n+1) \implies p|n^2+n+1$ or $p|n-1$. But, in the second case we get $p \le n-1$ and $n|p-1 \implies n \le p-1$, a contradiction. So, $n|p-1$ and $p|n^2+n+1$. Let $p-1=nm$. Then $m \le n+1$. If $m<n+1$, take $m=n-k$. $p=n(n-k)+1 \implies p|n(k+1) \implies p|k+1 \implies p =nm+1 \le n-m+1 \implies nm \le n-m$ But this is a contradiction. So, $p=n(n+1)+1 \implies 4p-3=(2n+1)^2$.

This is also Iran 2005 P1

djb86 wrote: An integer $n > 1$ and a prime $p$ are such that $n$ divides $p-1$, and $p$ divides $n^3 - 1$. Prove that $4p - 3$ is a perfect square. By $n|p-1$, we get $p=nk+1$ for some integer $k$. Hence $$nk+1|n^{2}+n+1 \Rightarrow nk+1 \le n^2+n+1 \Rightarrow k \le n+1 :(1)$$Now, since $gcd(k,nk+1)=1$, hence we get $$nk+1|kn^{2}+kn+k=n(nk+1)+(k-1)n+k \Leftrightarrow nk+1|(k-1)n+k$$$$\Rightarrow nk+1 \le (k-1)n+k \Rightarrow n+1 \leq k :(2)$$By (1) and (2), we get $k=n+1$. Thus, $p=n(n+1)+1=n^{2}+n+1 \Rightarrow 4p-3 = (2n+1)^{2}$, as desired.

since this is a very popular problem, I have seen it in at least 5 state olympiads.

this problem's solution in Justin Steven's Number theory