Find all integers $(x, y)$ such that $1 + 2^x + 2^{2x + 1} = y^2$ [duplicate]

Solution 1:

$1 + 2^x + 2^{2x + 1} = y^2 \\\implies 2^x(2^{x+1}+1) = (y+1)(y-1)$

$x<-1$ gives a non-integer LHS (no solutions)

$x=-1$ gives LHS $= 1$ with no solutions for $y$

$x=0$ gives LHS $= 3$ and $y=\pm 2$

For $x>0$, $y$ is odd so put $y=2k+1$ and $2^x(2^{x+1}+1) = (2k+2)(2k) = 4k(k+1)$ which is divisible by $8$ so $x\ge 3$ and $2^{x-2}(2^{x+1}+1) = k(k+1)$.

Clearly we cannot have $k=2^{x-2}$ or $k+1=2^{x-2}$ so we need $(2^{x+1}+1)$ to split into (odd) factors $r,s$ such that $2^{x-2}r = s\pm 1$.

Then $|r|<8$ otherwise $2^{x-2}|r| > (2^{x+1}+1)$. Also $2^{x-2}r^2 = sr\pm r$ so $|r|=3$ is the only viable choice and $2^{x-2}9 = (2^{x+1}+1) \pm 3$ gives $2^{x-2} = 4$ i.e. $x=4$ (and $y=\pm 23$) as the only other solution.

In summary: the only solutions $(x,y)$ are $(0,\pm 2)$ and $(4,\pm 23)$, those you found.