1. Determine all natural numbers $x, y$, and $z$ for which

$$
1+2^{x} 3^{y}=z^{2}
$$
Solution. It is easily verified that for $z=1,2,3$, the given equation has no solution.

Let $z \geq 4$. We have $2^{x} 3^{y}=(z-1)(z+1)$. At most one of $z-1$ and $z+1$ is divisible by 3, because if $3 \mid z-1$ and $3 \mid z+1$ then $3 \mid(z+1)-(z-1)=2$, which is impossible. Similarly, since $2 \mid(z-1)(z+1)$, both $z-1$ and $z+1$ are divisible by 2, and only one of them can be divisible by 4, because if $4 \mid z-1$ and $4 \mid z+1$ then $4 \mid(z+1)-(z-1)=2$, which is impossible. Therefore, we have the following cases:

$1^{\circ} z+1=2 \cdot 3^{y}, z-1=2^{x-1}$ and

$2^{\circ} z+1=2^{x-1}, z-1=2 \cdot 3^{y}$.

Case $1^{\circ}$ Let $z+1=2 \cdot 3^{y}, z-1=2^{x-1}$.

By subtracting the given equations, we have

$$
2 \cdot 3^{y}-2^{x-1}=2 \text { i.e. } 3^{y}-2^{x-2}=1
$$

For $x=2$, it is easily verified that the given equation has no solution.

For $x=3$, we have $3^{y}=1+2=3$ i.e. $y=1$ and it is easily obtained that $z=5$.

Thus, one solution is $(x, y, z)=(3,1,5)$.

If $x \geq 4$, we have $3^{y} \equiv 1(\bmod 4)$, so it follows that $y$ is even, i.e., $y=2 y_{1}$, $y_{1} \in \mathbb{N}$.

Now we have $3^{2 y_{1}}-1=2^{x-2}$ i.e. $\left(3^{y_{1}}-1\right)\left(3^{y_{1}}+1\right)=2^{x-2}$, from which it is easily obtained that $3^{y_{1}}-1=2$ and $3^{y_{1}}+1=2^{x-3}$, i.e., $y_{1}=1$ or $y=2$, and now it is easily obtained that $x=5$ and $z=17$.

Thus, the solution is $(x, y, z)=(5,2,17)$.

Case $2^{\circ}$ Let $z+1=2^{x-1}, z-1=2 \cdot 3^{y}$, then $2^{x-1}-2 \cdot 3^{y}=2$, i.e.

$$
2^{x-2}-3^{y}=1
$$

For $y=1$, we have $x=4$ and $z=7$, so the solution is $(x, y, z)=(4,1,7)$.

If $y \geq 2$, then we have $2^{x-2} \equiv 1(\bmod 3)$, i.e., $x-2$ must be an even number, so let $x-2=2 x_{1}$. Then we have $3^{y}=2^{2 x_{1}}-1=\left(2^{x_{1}}-1\right)\left(2^{x_{1}}+1\right)$, from which it follows that $2^{x_{1}}-1=1$ or 3. If $2^{x_{1}}-1=1$, we get $x_{1}=1$ i.e., $x=4$, and we obtain the solution $(4,1,7)$.

If $2^{x_{1}}-1=3$, then $x_{1}=2$ i.e., $x=6$ and we get $3^{y}=15$, which is impossible.

Thus, the solutions to the given equation are $(x, y, z)=(3,1,5),(5,2,17),(4,1,7)$.