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.

From your last expression we have $$\tag1(p+1)(p-1)=2\cdot[\;2(2d^2+2d)+1\;].$$  If $p=2$ then you get from the original expression that $2q^2=3,$ which is impossible. Therefore $p>2$ and hence $p$ is odd. Then $p-1$ and $p+1$ are both even, which implies that their product $(p+1)(p-1)$ is divisible by $4$ which is clearly false by $(1).$ Therefore $q$ must be even and thus it must be equal $2,$ which implies that $p=3.$
Another approach: Since $p^2-2q^2=1$ then $(p+1)(p-1)=2q^2$ and the Unique Factorization Theorem implies that either $p+1=q^2\;\wedge\;p-1=2$ or $p+1=2q\;\wedge\;p-1=q$ but in either case $p=3$ and $q=2.$
