determin all pair (x,y) of integer such that 1+2^x+2^(2x+1)=y^2

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2^x ( 2^(x+1) +1 )=(y+1)(y-1). Now y must be odd (unless x=0). Since 2 must strictly divide exactly one of y-1 and y+1 we have that 2^(x-1) strictly divides the other factor.
Case I) 2 || y-1. Then y-1=2k and y+1=2^(x-1) j for some odd integers k and j. Plugging these into the equation we have (*) 2^(x+1) + 1 = kj. On the other hand, since y-1=2k and y+1=2^(x-1) j we have 2k+1=2^(x-1)j - 1 so that (**) k+1=2^(x-2) j. Substituting 2^(x-2) from (**) into (*) gives k = (j+8)/(j^2 - 8). This is an integer for odd j only if j=1 or 3. Of these, only j=3 produces integer k, namely k = 11. Then by plugging j=3 and k=11 into (**) we obtain x=4.
Case II) 2|| y+1. Then y+1=2k and y-1=2^(x-1) j for some odd integrs k and j so that again (*) 2^(x+1)+1=kj. Solving for y in each, equating, and simplifying yields (***) k-1=2^(x-2) j. Solving for 2^(x-1) in (***) and plugging into (*) we have k=(j-8)/(j^2-8); an integer only if j=1 or 3 since j is odd. If j=3 then k is negative, hence j=1 so that k=1. But then by (***) 2^(x-2)=0 which is absurd.
Case III) x=0. Then y^2=1+1+2=4 so y=2 or -2.
Hence (4,23), (4,-23), (0,2) and (0,-2) are the only integer solutions.