Solve equation in prime numbers 
Solve the equation in prime number $$p^3+q^3+1=p^2 q^2.$$

I have found the solutions $(2,3), (3,2)$ and need to prove that there are no other solutions. I think that there is an inequality $p^3+q^3+1\leq p^2 q^2$ for $p,q >2 $  but  how to prove it.

My solution
Case 1. Let $p \mod 3=q \mod 3=1$. Then $p^3+q^3+1=3=0  \mod 3 \neq p^2 q^2=1 \mod 3$
Case 2 $p \mod 3=q \mod 3=2$. Then $p^3+q^3+1=2+2+1=2  \mod 3 \neq p^2 q^2=1 \mod 3$
Сase 3. $p \mod 3=1, q \mod 3=2$. Then $p^3+q^3+1=1+2+1=1  \mod 3 \neq p^2 q^2=1 \cdot 2 =2 \mod 3$
So, must be, say $p=0 \mod 3 \implies p=3$.  We get the equation $q^3+28=9 q^2 $ or $q^3-9 q^2+28=\left( q-2 \right)  \left( {q}^{2}-7\,q-14 \right) $ and $q=2.$ By symmetry another solution is $p=2, q=3.$
