Example $6([47.4])$ Find all integer pairs $\{x, y\}$ such that $1+2^{x}+2^{2 x+1}=y^{2}$.
Solve: First, exclude simple cases. $x$ cannot be a negative integer (why). When $x=0$, $y=\pm 1$. Below, we discuss the case where $x \geqslant 1$.
(i) $y$ must be an odd number, so $y^{2}$ is a positive integer of the form $8k+1$. This implies $x \geqslant 3$ (why). We will assume $y$ is positive for now.
(ii) Rewrite the equation as
$$2^{x}\left(1+2^{x+1}\right)=(y-1)(y+1)$$

which is equivalent to
$$2^{x-2}\left(1+2^{x+1}\right)=(y-1) / 2 \cdot (y+1) / 2.$$
Since the positive integers $(y-1) / 2$ and $(y+1) / 2$ differ by 1, one must be odd and the other even.
(iii) If $(y-1) / 2$ is even, then $2^{x-2}$ divides $(y-1) / 2$ (why), so we can set $(y-1) / 2=2^{x-2} n$, where $n$ is a positive odd number. Therefore, $(y+1) / 2=2^{x-2} n+1$. From (ii), we have
$$2^{x-2} n\left(2^{x-2} n+1\right)=2^{x-2}\left(1+2^{x+1}\right)$$

which simplifies to
$$n\left(2^{x-2} n+1\right)=1+2^{x+1}$$

Thus,
$$n-1=2^{x-2}\left(8-n^{2}\right)$$

This equation cannot hold, as it requires $8-n^{2} \geqslant 0$ (why). Therefore, $n$ must be 1 or 2, neither of which satisfies the equation.
(iv) If $(y+1) / 2$ is odd, then $2^{x-2}$ divides $(y+1) / 2$ (why), so we can set $(y+1) / 2=2^{x-2} n$, where $n$ is a positive odd number. Therefore, $(y-1) / 2=2^{x-2} n-1$. From (ii), we have
$$2^{x-2} n\left(2^{x-2} n-1\right)=2^{x-2}\left(1+2^{x+1}\right)$$

which simplifies to
$$n\left(2^{x-2} n-1\right)=1+2^{x+1}$$

Thus,
$$n+1=2^{x-2}\left(n^{2}-8\right)$$

Since $2^{x-2}\left(n^{2}-8\right) \geqslant 2\left(n^{2}-8\right)>0$, $n$ can only be 3 (why), giving
$$x=4, \quad y=23$$
(v) Combining all the above discussions, the complete solutions are $x=0, y= \pm 2 ; x=4, y= \pm 23$.