IMO 2006 Shortlisted Problems

If $(x,y)$ is a solution then obviously $x\ge 0$ and $(x,-y)$ is a solution too. For $x=0$ we get the two solutions $(0,2)$ and $(0,-2)$.

Now let $(x,y)$ be a solution with $x>0$; without loss of generality confine attention to $y>0$. The equation rewritten as $$2^x\left(1+2^{x+1}\right)=(y-1)(y+1)$$ shows that the factors $y-1$ and $y+1$ are even, exactly one of them divisible by $4$. Hence $x\ge 3$ and one of these factors is divisible by $2^{x-1}$ but not by $2^x$. So $$y=2^{x-1}m+ϵ,m\text{ odd},ϵ=\pm 1$$ Plugging this into the original equation we obtain $$2^x\left(1+2^{x+1}\right)={\left(2^{x-1}m+ϵ\right)}^2-1=2^{2x-2}{m}^2+2^{x}mϵ$$ or, equivalently $$1+2^{x+1}=2^{x-2}{m}^2+mϵ$$ Therefore $$1-ϵm=2^{x-2}\left(m^2-8\right)$$ For $ϵ=1$ this yields $m^2-8\le 0$, i.e., $m=1$, which fails to satisfy the previous equation.

For $ϵ=-1$ equation gives us $$1+m=2^{x-2}\left(m^2-8\right)\ge 2\left(m^2-8\right)$$ implying $2m^2-m-17\le 0$. Hence $m\le 3$; on the other hand $m$ cannot be $1$ by the previous equation. Because $m$ is odd, we obtain $m=3$, leading to $x=4$. From the previous formula we get $y=23$. These values indeed satisfy the given equation. Recall that then $y=-23$ is also good. Thus we have the complete list of solutions $(x,y):(0,2),(0,-2),(4,23),(4,-23)$.