I claim $(p,q)=(2,5)$ is the only solution. Note that if $p=q$ then $p^4+p^2+10p=p^2+3$, which doesn't have a solution. So, $p\ne q$. Using $q\mid p(p^4+p^2+10q)$, we find $q\mid p^2(p^2+1)$, so $q\mid p^2+1$ as $q\ne p$ are primes. Likewise, $p\mid q^2+3$. Now if $p=2$ then $q=5$, which is easily seen to work. So, $p\ge 3$. Notice also that $q\ne 3$ (as $2$ is not a quadratic residue modulo 3), and $q=2$ yields $p=7$ which doesn't work. So, $q\ge 5$. Using $p\mid q^2+3$, we thus find $(-3/p)=1$, which means $p\equiv 1\pmod{6}$ (a well-known fact). So, $p(p^4+p^2+10q)\equiv 4q+2$, yielding $q^3\equiv q+2\pmod{3}$. This is a contradiction as $q^3\equiv q\pmod{3}$.

If $p=5$ then $q$ has no solution. Assume $q$ is not equal to 5. We can take $\pmod{5}$ then $LHS=p(p^4+p^2+10q)\equiv p(1+p^2) \equiv 2,0,3 \pmod{5}$ and $RHS=q(q^2+3) \equiv 4,1 \pmod{5}$,so we have the only solution $(p,q)=(2,5)$.

$1st$ case, $p$ even Since it's a prime then $p=2 \implies 2(20+10q)=q(q^2+3)$ so $q^3-17q-40=0$ we easily see that $7^3-17*7>40$ so $q\le 5$ and in all cases only $q=5$ works. $2nd$ case, $q$ even $q=2 \implies p(p^4+p^2+20)=14$, a contradiction. $3rd$ case, $p$ and $q$ odd Notice that $q^2+3$ will be even so $p(p^4+p^2+10q)$ is even but $p^4+p^2+10q$ is odd $\implies p=2$, contradiction. With this we have that $(p,q)=(2,5)$

https://artofproblemsolving.com/community/c6h3038033p27345873

Obviusly $q^2 \equiv 3(mod 4)$ so $4\mid q(q^2+3)$ so $4\mid p(p^4+p^2+10q)$ if $p$ is odd so right hand is odd this is a contradiction so $p = 2$, and then $q =5$