There are no prime number $p$ and $q$ such that $p^2(p^3-1)=q(q+1)$. Suppose in the sake of contradiction that there exist prime numbers $p,q$ that satisfy the given equation. First of all,let's see that $p\neq q$.So $(p,q)=1$.Thus,since $p^2|q(q+1)$,we obtain that $p^2|q+1$.Let $q=kp^2-1,k\in\mathbb{N*}$. Since $(p,q)=1$ and $q|p^2(p^3-1)$,we obtain that $q|p^3-1=(p-1)(p^2+p+1)$.Observe that $(p-1,p^2+p+1)\in \{1,3\}$ and $q=3$ is not a solution,so $q|p-1$ or $q|p^2+p+1$.Obviously $q|p-1$ is impossible(since $q|p-1$ implies $q\le p-1$,but $p^2|q+1$ implies $q\ge p^2-1$),thus $q|p^2+p+1$. This implies $kp^2-1=q\le p^2+p+1$.Suppose that $k\ge 2$.Then $kp^2-1\ge 2p^2-1>p^2+p+1$(because it's equivalent to $p^2>p+2$ and $p^2=p+2$ is true only for $p=2$,but $p=2$ is not a solution). Therefore $k=1$.We have $q=p^2-1=(p-1)(p+1)$,and since $p>2$ we would obtain that $q$ is not a prime number,contradiction!

This equality is p^5-1=q^2+p^2, and it is known (it can be used as if it was a theorem, or proofed and called a lemma) that all prime divisors bigger than 3 of a sum of two squares are 1 modulo 4. The left side is (p-1)(p^4+p^3+p^2+p+1), so if p>3, p is 2 modulo 4, which is a contradiction. So we just need to check if p=2 or p=3 works. For p=2, we get q^2=27, and for p=3 we get q^2=233, and neither 27 nor 233 are full squares, so we have a contradiction, there are no such numbers

Rewrite it as $p^2(p-1)(p^2+p+1)=q(q+1)$. Clearly $p\neq 2$. Also it is clear $q>p$ and therefore $q $ is odd. 1. Dividing by $p^2$, $p^2|q+1$, so $2p^2|q+1$ since $q+1$ is even. 2. Dividing by $q$, $q | p^2+p+1$. From 1. and 2. We have $q < p^2+p+2<2p^2 \le q+1$ , contradiction, since there is no integers between $q$ and $ q+1$.

We have that $q | p^3-1=(p-1)(p^2+p+1)$ so we have two cases: $1)$ $q|p-1$. Thus $p=kq+1$ for a positive integer $k$, but we have that $p^2|q+1$, which implies $q+1\geq(kq+1)^2$, contradiction. $2)$ $q|p^2+p+1$. Now we have $\frac{p^2+p+1}{q}=\frac{q+1}{p^2(p-1)}$ and thus $p^3-p^2\leq q+1\leq p^2+p+2$, which rearranges to $(p-1)(p+1)(p-2)\leq4$, which is not satisfied by any prime $p$. $\blacksquare$

Just prove that it must be q+1 | p^2 and we are done since q+1 is even so p=2 .... No solutions!