[Number Theory]
Problem: Determine all pairs $(x, y)$ of integers satisfying the equation

$$
1+2^{x}+2^{2 x+1}=y^{2}
$$

Solution: If $(x, y)$ is a solution then obviously $x \geq 0$ 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 $x \geq 3$ and one of these factors is divisible by $2^{x-1}$ but not by $2^{x}$. So

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

Plugging this into the original equation we obtain

$$
2^{x}\left(1+2^{x+1}\right)=\left(2^{x-1} m+\epsilon\right)^{2}-1=2^{2 x-2} m^{2}+2^{x} m \epsilon,
$$

or, equivalently

$$
1+2^{x+1}=2^{x-2} m^{2}+m \epsilon .
$$

Therefore

$$
1-\epsilon m=2^{x-2}\left(m^{2}-8\right) .\tag{2}
$$

For $\epsilon=1$ this yields $m^{2}-8 \leq 0$, i.e., $m=1$, which fails to satisfy (2).

For $\epsilon=-1$ equation (2) gives us

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

implying $2 m^{2}-m-17 \leq 0$. Hence $m \leq 3$; on the other hand $m$ cannot be 1 by $(2)$. Because $m$ is odd, we obtain $m=3$, leading to $x=4$. From (1) 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 $(x, y):(0,2),(0,-2),(4,23),(4,-23)$.

Final answer: $(0,2),(0,-2),(4,23),(4,-23)$