Find all primes $(p,q)$ such that $pq$ divides $p^3 + q^3 +1$ The question is

Find all primes $(p,q)$ such that $pq$ divides $p^3 + q^3 +1$.

My attempt:
This reduces to finding primes $p,q$ such that p divides $q^3+1=(q+1)(q^2-q+1)$ and q divides $p^3+1=(p+1)(p^2-p+1)$.
Now we have 4 cases: If p divides $q+1$ and q divides $p+1$, then $\{p,q\}=\{2,3\}$.
I don't know how to deal with the other cases. For example, what happens if p divides $q+1$ and q divides $p^2-p+1$?

If $p \mid q+1$ and $q \mid p^2-p+1$, then $\frac{p^2-p+1}q$ must be $-1$ modulo $p$, since $q\equiv -1\pmod p$. As a result, there exist positive $m,n$ for which
$$(mp-1)(np-1)=p^2-p+1.$$
We clearly can't have $m=n=1$, so we must have
$$(p-1)(2p-1)\leq (mp-1)(np-1)=p^2-p+1\implies p\leq 2,$$
and thus $(2,3)$ is the only solution here as well. The same goes for if $p$ and $q$ are swapped.
For the final case, we claim that $a \mid b^2-b+1$ and $b \mid a^2-a+1$ has no solutions in positive integers besides $(a,b)=(1,1)$. To show this, assume for the sake of contradiction that there exists such a pair with minimal sum, and without loss of generality assume $a>b$. Now, consider
$$c=\frac{b^2-b+1}a.$$
Since $b>1$ (as otherwise $a=b=1$), $c<a$, so $b+c<a+b$. However, $c \mid b^2-b+1$, and
$$c^2-c+1\equiv \frac1{a^2}-\frac1a+1\equiv \frac{a^2-a+1}{a^2}\equiv 0\pmod b$$
(using that $\gcd(a,b)=1$), so $b \mid c^2-c+1$ as well. Then $(b,c)$ is a pair satisfying the conditions with smaller sum than $(a,b)$, a contradiction. So, not only can there not exist primes $(p,q)$ with $p \mid q^2-q+1$ and $q \mid p^2-p+1$, there can't exist any such pairs of positive integers.
