885. Solve in prime numbers $x, y$ and $z$ the equation

$$
x^{y}+1=z
$$
$\triangle$ The number $z$ is greater than 2, because if $z=2$, then $x=1$, which is impossible. Therefore, $z$ is odd, and consequently, the number $x$ is even. But $x$ is a prime number, so $x=2$. We get the equation:

$$
2^{y}+1=z
$$

Is the number $y$ even or odd? If $y$ is odd, then the sum $2^{y}+1$ is divisible by 3 (see § 7, third divisibility), and the quotient of such division is greater than 1; but in this case, the number $z$ is composite. Therefore, the number $y$ is even, i.e., $y=2$. Let's calculate $z$: $z=2^{2}+1=5$.

Answer: $(2 ; 2 ; 5)$.