Determine all pairs $(x, y)$ of natural integers such that $2^{2 x+1}+2^{x}+1=y^{2}$
We notice that $x=0, y=2$ is a solution. From now on, assume $x \geqslant 1$. We start by factoring: $2^{2 x+1}+2^{x}=2^{x}\left(1+2^{x+1}\right)=y^{2}-1=(y-1)(y+1)$. Since $x \geqslant 1, (y+1)(y-1)$ is even, so $y$ is odd. Thus, both $y-1$ and $y+1$ are even, but since their difference is 2, only one is divisible by 4. In particular, either $y+1$ or $y-1$ is divisible by $2^{x-1}$.

- If $2^{x-1}$ divides $y-1$, we set $y=1+2^{x-1} \times k$ for $k$ a natural number. We substitute back into the initial equation: $2^{2 x+1}+2^{x}+1=y^{2}=1+2^{x} k+2^{2(x-1)} k^{2}$. We get $2^{x+1}+1=k+k^{2} 2^{x-2}$, which simplifies to $1-k=2^{x-2}\left(k^{2}-8\right)$. Looking modulo 2, we see that $k$ is odd. If $k \geqslant 1$, $1-k$ is negative, so $k^{2}-8 \leqslant 0$ i.e., $0 \leqslant k \leqslant 2$. By parity, $k=1$, which contradicts $1-k=2^{x-2}\left(k^{2}-8\right)$.
- If $2^{x-1}$ divides $y+1$, we set $y=-1+2^{x-1} \times k$ for $k$ a natural number. We substitute back into the initial equation: $2^{2 x+1}+2^{x}+1=y^{2}=1-2^{x} k+2^{2(x-1)} k^{2}$. We get $2^{x+1}+1=-k+k^{2} 2^{x-2}$, which simplifies to $1+k=2^{x-2}\left(k^{2}-8\right)$. Looking modulo 2, we see that $k$ is odd. Since $1+k$ is positive, $k^{2}-8$ is also positive, so $k \geqslant 3$. We have $1+k=2^{x-2}\left(k^{2}-8\right) \geqslant 2 k^{2}-16$, which rearranges to $2 k^{2}-k-17 \leqslant 0$. For $k \geqslant 4, 2 k^{2}-k-17=k(2 k-1)-17 \geqslant 4 \times 7-17=11>0$, so $3 \leqslant k \leqslant 4$. Since $k$ is odd, $k=3$, thus $4=2^{x-2}(9-8)=2^{x-2}$, so $x=4$. We also have $y=-1+3 \times 2^{4-1}=23$.

Thus, we have found that all solutions are of the form $(4,23)$ and $(0,2)$. It remains to verify that these two pairs are indeed solutions. Conversely, since $2^{1}+2^{0}+1=4=2^{2}$ and $2^{9}+2^{4}+1=512+16+1=529=23^{2}$, $(4,23)$ and $(0,2)$ are the two solution pairs.