Can't even use the perfect square identity!

Firstly, observe that if $(x, y)$ is a solution, then obviously $x \geq 0$ and $(x, -y)$ is another solution. For $x=0$ , we get 2 solutions: $(0,2)$ and $(0, -2)$ .

WLOG $x>0, y>0$ . Thus, the original statement can be rewritten as

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

which implies that $y-1$ and $y+1$ are both even, so one of them must be divisible by 4. Hence, $x \geq 3$ , and one of these factors is divisible by $2^{x-1}$ but not $2^x$ . So

$y=2^{x-1}m+\epsilon, \quad m \text{ odd}, \quad \epsilon=\pm 1$

Plugging this into the original equation, we get

$\begin{aligned} 2^x (1+2^{x+1}) &= (2^{x-1}m+\epsilon)^2 - 1\\ &= 2^{2x-2} m^2 + 2^xm\epsilon\\ 1+2^{x+1} &= 2^{x-2} m^2 + m \epsilon\\ 1-m\epsilon &= 2^{x-2} (m^2 - 8) \end{aligned}$

For $\epsilon = 1$ , we get $m^2-8\leq 0$ , so $m=1$ , which doesn't satisfy this equation. If $\epsilon = -1$ , we have

$1+m = 2^{x-2} (m^2 - 8) \geq 2(m^2 - 8)$

which implies $2m^2 - m - 17 \leq 0$ . Hence, $m \leq 3$ . However, we already know $m \neq 1$ . Thus, $m=3$ , so $x=4$ . These values indeed satisfy our equation, and we get $y=\pm 23$ . Therefore, there are 4 solutions: $(0, 2)$ , $(0, -2)$ , $(4, 23)$ and $(4, -23)$ .