Find all primes $p$ and $q$ such that $p^2-2q^2=1.$ 
Find all primes $p$ and $q$ such that $p^2-2q^2=1.$

My idea so far was to observe that since $2q^2$ is even, then $q^2$ must be odd or even. If $q^2$ is even, then $q$ is even and the only even prime is $2.$ Thus one pair of primes $(p,q)=(3,2).$
But, if $q^2$ is odd, then $q$ is also odd and $q=2d+1$ for some integer $d.$ Then,
$p^2-2q^2=1$ turns into $p^2=4(2d^2+2d)+3.$
From here I do not know how to proceed.
Any ideas or suggestions would be greatly appreciated.

$p^2-2q^2\equiv p^2+q^2\equiv 1\pmod{3}$.
But $a^2\equiv \{0,1\}\pmod{3}$ for all $a\in\Bbb Z$, so exactly one of $p,q$ is equal to $3$. 
$q=3$ gives $p^2=19$, impossible. $p=3$ gives $(p,q)=(3,2)$.
