124. Solve the equation: $x^{y}+1=z$ in prime numbers.
124. One of the numbers $x^{y}$ and $z$ is even, and the other is odd. If $z$ is even, then being prime, it equals 2; but then $x$ and $y$ can only equal one or $x$ can be any number and $y$ equals zero; therefore, $x$ and $y$ will not be prime numbers. It remains to assume that $x^{y}$ is even, then $x=2$ and $2^{y}+1=z$. If $y$ is odd, then $2^{y}+1$ is divisible by $2+1$, i.e., it will be a composite number, therefore, $y$ is even, so $y=2$, hence $z=2^{2}+$ $+1=5$.

Answer. $2 ; 2 ; 5$.