3. In the set of integers, solve the equation: $x^{y}-2^{z}=1$.
Solution. Let $(x, y, z)$ be an ordered triplet of integers for which $x^{y}-2^{z}=1$. It is easily seen that it is not possible for $y \leq 0$ and that for $y=1$ every triplet

$$
\left(2^{z}+1,1, z\right), \text { where } z \in\{0,1,2, \ldots\}
$$

is a solution to the given equation. Let $y>1$. Then $|x|>1, z>1$, and moreover, $x$ is an odd number and it holds that

$$
(x-1)\left(x^{y-1}+x^{y-2}+\ldots+x+1\right)=2^{z}
$$

Now, since $x^{y-1}+x^{y-2}+\ldots+x+1$ is an even number, and all the terms are odd, we conclude that $y-1=2 k+1$, where $k$ is a non-negative integer, so

$$
(x-1)(x+1)\left(x^{y-2}+x^{y-4}+\ldots+1\right)=2^{z}
$$

The number $(x-1)(x+1)$ is a power of the number 2 with a natural exponent if and only if $x \in\{-3,3\}$. For $x=3$ we get

$$
\begin{aligned}
& 9^{k}+9^{k-1}+\ldots+1=2^{z-3} \\
& 2^{z}=9^{k+1}-1=\left(3^{k+1}-1\right)\left(3^{k+1}+1\right)
\end{aligned}
$$

from which we get $k=0, y=2, z=3$, i.e., the solution is the triplet $(3,2,3)$. For $x=-3$ we get the solution $(-3,2,3)$.