Diophantine equation involving prime numbers : $p^3 - q^5 = (p+q)^2$

$$p^3-q^5=(p+q)^2=p^2+2pq+q^2$$ So, $p^3-q^5\ge 0$, so $p^3\ge q^5$, so $p>q$. $$p^2(p-1)=p^3-p^2=q^5+2pq+q^2=q(q^4+2p+q)$$

Hence, $p^2$ divides $q^4+2p+q$ (as it can't divide q) and so $q$ divides (p-1).

So $p^2\le q^4+2p+q$, so $p\le q^2+1$. We can then verify easily that $q=2$ has no solution for $p$.

But $p$ is prime, and $q^2+1$ and $q^2$ are not so : $$q<p\le q^2-1$$

So let $p=aq+b$ with $1\le a< q$ and $1\le b<q$.

We have $$q^5=p^3-(p+q)^2=(b^3-b^2)+q.(\dots)$$ Hence $q$ divides $b^2(b-1)$, so $b=1$

Hence $p=aq+1$ with $1\le a< q$

We have $$q^5=p^3-(p+q)^2=q.(3a-2(a+1))+q^2.(\dots)$$

Hence, $q$ divides $a-2$, so $a=2$.

So $p=2q+1$

$$q^5=q^2.(8q+12-9)$$ $$0=q^3-8q-3=(q-3).(q^2+3q+1)$$

The only positive integer solution is $q=3$ and so $p=7$.

At least one of $p,q$ is $3$. If not, then by little Fermat $p^2\equiv q^2\equiv 1,\, p^3\equiv p,\, q^5\equiv q\pmod{\! 3}$, so

$$p^3-q^5\equiv (p+q)^2\iff p-q\equiv 2+2pq$$

$$\iff pq+p-q+1\equiv 0\iff (p-1)(q+1)\equiv 1\pmod{\! 3}$$

For LHS to be non-zero, $p\equiv -1,\, q\equiv 1\pmod{\! 3}$, but still congruence doesn't hold.

$p^3>q^5$, so $p\neq 3$. Then $q=3$ and $p\ge 7$. $$p^3-p^2-6p-(3^5+9)=0,$$

so $p\mid 3^5+9\iff p\mid 3^3+1=28\iff p=7$; $\,(p,q)=(7,3)$ works.