A Diophantine equation

Let us determine all the prime positive p and q such that p²-2q²=1. We observe that p=3 and q=2 satisfy the question since they are both positive primes and (3)²-2(2)²=9-8=1. Conversely, if p and q are prime positives such that p²-2q²=1 then p²=2q²+1 therefore p² is odd or p is odd let’s say p=2k+1 then substituting in the equation we find 4k²+4k+1=2q²+1 it follows that q²=2k²+2k i.e. q²=2v with v integers it follows that q² is even i.e. q is even therefore q=2 from which p²=9 and therefore p=3. Therefore: {(p,q) ∈ ℤ² : p²-2q²=1 , p,q prime positives}={(3,2)}