4. Solve the equation $2^{x}+3^{y}+1=6^{z}$ in natural numbers.


Answer: $(1,1,1),(3,3,2),(5,1,2)$.

Solution. The right-hand side of the equation is divisible by 3, and the left-hand side gives the same remainder when divided by 3 as $(-1)^{x}+1$. Therefore, $x$ is odd. Let's consider two possible cases.

1) $x=1$. Note that either $y=z=1$, or $y \geqslant 2$ and $z \geqslant 2$. The first case satisfies the equation: $2^{1}+3^{1}+1=6^{1}$, while the second is impossible because the right-hand side is divisible by 9, but the left-hand side is not.
2) $x \geqslant 3$. Note that $6^{z} \geqslant 2^{3}+3+1=12$, hence $z \geqslant 2$. Then $2^{x}$ and $6^{z}$ are divisible by 4, while $3^{y}$ gives the same remainder when divided by 4 as $(-1)^{y}$. Therefore, $y$ is odd, i.e., $y=2 n+1$. The left-hand side of the equation is not divisible by 8, since

$$
2^{x}+3^{y}+1=2^{x}+3^{2 n+1}-3+4=2^{x}+3\left(3^{n}-1\right)\left(3^{n}+1\right)+4
$$

and $\left(3^{n}-1\right)\left(3^{n}+1\right)$ is divisible by 8 as the product of two consecutive even numbers. Then $6^{z}$ is also not divisible by 8, which is only possible when $z=2$. It remains to solve the equation $2^{x}+3^{y}=35$. For $x$, only the values 3 and 5 are possible, and they correspond to $y=3$ and $y=1$.