+ Find all integer pairs $(x, y)$ such that
$$
1+2^{x}+2^{2 x+1}=y^{2} .
$$
If $(x, y)$ is a solution, then $x \geqslant 0, (x, -y)$ is also a solution. When $x=0$, the solutions are $(0,2), (0,-2)$.
Assume $(x, y)$ is a solution, $x>0$, without loss of generality, let $y>0$.
The original equation is equivalent to
$$
2^{x}\left(1+2^{x+1}\right)=(y-1)(y+1).
$$

Thus, $y-1$ and $y+1$ are even numbers, with exactly one being divisible by 4. Therefore, $x \geqslant 3$, and one of the factors is divisible by $2^{x-1}$ but not by $2^{x}$. Hence, $y=2^{x-1} m+\varepsilon, m$ is an odd number, $\varepsilon= \pm 1$. Substituting into the original equation, we get
$$
2^{x}\left(1+2^{x+1}\right)=\left(2^{x-1} m+\varepsilon\right)^{2}-1=2^{2 x-2} m^{2}+2^{x} m \varepsilon,
$$

which simplifies to
$$
1+2^{x+1}=2^{x-2} m^{2}+m \varepsilon,
$$

thus,
$$
1-m \varepsilon=2^{x-2}\left(m^{2}-8\right).
$$

When $\varepsilon=1$, $m^{2}-8 \leqslant 0$, i.e., $m=1$, the above equation does not hold.
When $\varepsilon=-1$, we have $1+m=2^{x-2}\left(m^{2}-8\right) \geqslant 2\left(m^{2}-8\right)$, which implies $2 m^{2}-m-17 \leqslant 0$, hence, $m \leqslant 3$. On the other hand, $m \neq 1$, since $m$ is an odd number, we get $m=3$, thus $x=4, y=23$.
Therefore, all solutions are $(0,2), (0,-2), (4,23), (4,-23)$.