891. Solve the equation in natural numbers

$$
3^{x}-2^{y}=1
$$
$\triangle$ First, let's determine whether the numbers $x$ and $y$ are even or odd.

If we write the equation as:

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

it is not hard to guess that the number $y$ is odd: only when $y$ is odd does the sum $2^{y}+1$ divide by the sum of the bases of the powers, i.e., by 3 (see § 7, second divisibility).

Now let's represent the equation as $2^{y}=3^{x}-1$, factoring the right-hand side (see § 7, identity (2)):

$$
2^{y}=2 \cdot\left(3^{x-1}+3^{x-2}+\ldots+3+1\right)
$$

Let's recall that the number of terms in the parentheses on the right-hand side of the last equation is equal to the exponent, i.e., $x$.

If $y>1$, then after dividing the equation by 2, its left-hand side, equal to $2^{y-1}$, remains even. Then the right-hand side is also even, which means $x$ is even.

And what if $y=1$? We need to consider two cases.

1) Suppose $y=1$. In this case, from the original equation we get:

$$
3^{x}-2=1, \quad 3^{x}=3, \quad x=1
$$

2) Suppose $y>1$. Then, as proven above, $x$ is even: $x=2 n(n \in N)$. Let's write the equation as:

$$
3^{x}-1=2^{y}, \quad 3^{2 n}-1=2^{y}, \quad\left(3^{n}+1\right)\left(3^{n}-1\right)=2^{y} .
$$

Thus, each of the numbers $3^{n}+1$ and $3^{n}-1$ is a power of two. But the greatest common divisor of these numbers is 2 (verify this), so the last equality holds only in one case, when

$$
\left\{\begin{array}{l}
3^{n}+1=2^{k} \\
3^{n}-1=2
\end{array}\right.
$$

where $k$ is a natural number greater than 1. From the second equation of the system, we find $n$: $n=1$. Then from the first equation, we find $k$:

$$
3+1=2^{k}, \quad 2^{k}=4=2^{2}, \quad k=2
$$

It remains to calculate $x$ and $y$:

$$
x=2 n=2, \quad y=k+1=3 .
$$

Answer: $(1 ; 1),(2 ; 3)$