Find all positive prime numbers $p$ and $q$ such that $p^{2}-q^{3}=1$.

Remark. $p$ is prime if it has only two divisors: 1 and itself. The numbers $2,3,5,7,11,13$ are prime, but $1,4,6,8,9$ are not.
If both of the numbers $p$ and $q$ are odd, then each of $p^{2}$ and $q^{3}$ has to be odd so their difference must be even. Hence, at least one of $p$ and $q$ has to be even. Since 2 is the only even prime number, and at the same time it is the smallest prime number, we must have $q=2$. Now we get $p=3$.