For $p=3$ we get $q=17$ $p^3(p^2+1)=(q-2)(q+1)$ Because $(q-2,q+1)={1,3}$ So because $p$ is not equal to $3$ devide only one of ${q-2,q+1}$ . Exactly the same for $p^3$ Because $q-2<q+1$ then $p^3=q+1$ $p^2+1=q-2$ $p^3-p^2-4=0$ so only solution for $p=2$ and $q=7$

To be more precise, for $p>3$ and the equivalent form $p^3(p^2+1)=(q-2)(q+1)$ it follows $p^3 \leq q+1$ and $p^2+1\geq q-2$, thus $p^3-p^2-4\leq 0$, impossible. Notice only the primality of $p$ has been used (like very often in such problems).

(p,q)=(2,7) and (3,17)

Lemma Be $x$ a positive integer, $$x(x-1)\equiv 0\mod{ 2}$$the proof is very easy So $$q^2-q\equiv 0 \mod{2}\Rightarrow p^5+p^3+2\equiv 0 \mod{2}$$$p\equiv 0 \mod{2}\Rightarrow p=2$ Now $$p^5+p^3+2=42=q^2-q=q(q-1)\rightarrow q=7$$ the answer is $(p,q)=(2,7)$

dkshield wrote: Lemma So $$q^2-q\equiv 0 \mod{2}\Rightarrow p^5+p^3+2\equiv 0 \mod{2}$$ you get nowhere with that since $p$ can also be $\equiv 1\pmod{2}$

Leicich wrote: Find all pairs of positive prime numbers $(p,q)$ such that $p^5+p^3+2=q^2-q$ $p^3(p^2+1)=(q-2)(q+1)$ If $p>3$ $\Rightarrow p^3|q-2$ or $p^3|q+1$ $\Rightarrow p^3\le q+1$ $\Rightarrow (q-2)(q+1)\le (q+1)(\frac{q+1}{p}+1)$ $\Rightarrow p(q-1)\le q+1$ As $p>3$ $\Rightarrow 3q-3<q+1$ $\Rightarrow q<2 (\Rightarrow \Leftarrow)$ If $p=3 \Rightarrow q=17$ If $p=2 \Rightarrow q=7$ $\Rightarrow (3,17)$ and $(2,7)$ are the only solutions$_\blacksquare$