2017/025) Solve $2^y+1 = x^2$

We have
$2^y = (x^2-1)= (x+1)(x-1)$
both x+ 1 and x-1 must be power of 2. difference of them is 2.
2 powers of 2 that differ by 2 are 2 and 4

so x+ 1= 4 and x-1 = 2 or x =3 and y = 3