What are all the solutions to
$$2^{2x}+2^{2y}+1=n^2 $$
I tried using the parametrization of Pythagorean Quadruples, but it did not work quite well.
There are $2$ parametrizations:
$(2np,2mp,p^2-n^2-m^2,p^2+n^2+m^2)$
Or $(mp+nq,np-mq,p^2+q^2-n^2-m^2,p^2+q^2+n^2+m^2)$
The problem is that the first one doesn't generate all solutions, but worked in my proof.So...any help?
EDIT:What we want to show is that the only solutions are
${(2^{2y-1}+1)}^2=4^{2y-1}+4^y+1$
We must have $$ 4^x + 4^y = (n-1)(n+1) \tag{1}$$ and $\gcd(n-1,n+1)\leq 2$, so by assuming $x\geq y$ and rewriting $(1)$ as $$ 4^y\cdot\left(4^{x-y}+1\right) = (n-1)(n+1) \tag{2}$$ there are just a few cases to check.