12・115 Find the prime solutions of the equation $2^{x+1}+y^{2}=z^{2}$.
(18th All-Russian Mathematical Olympiad, 1992)
[Solution] From the given equation, we have
$$
2^{x+1}=(z-y)(z+y),
$$

which implies
$$
\left\{\begin{array}{l}
z-y=2^{k} \\
z+y=2^{x-k+1} .
\end{array}\right.
$$

where $k \in \mathbb{Z}$, and $0 \leqslant k < x-k+1$.
Adding the two equations, we get
$$
z=2^{k-1}+2^{x-k} \text{. }
$$

Subtracting the two equations, we get
$$
y=2^{x-k}-2^{k-1} \text{. }
$$

If $k-1 \geqslant 1$, then $y$ and $z$ cannot both be prime numbers.
If $k=1$, then
$$
\begin{array}{l}
y=2^{x-1}-1 \\
z=2^{x-1}+1
\end{array}
$$

Since $y=2^{x-1}-1, 2^{x-1}$, and $z=2^{x-1}+1$ are three consecutive natural numbers, one of them must be divisible by 3. Since $2^{x-1}$ is not divisible by 3, either $z$ or $y$ must be divisible by 3.
If $z$ is divisible by 3, then $z=3$, in which case $y=1$, and $y$ is not a prime number.
If $y$ is divisible by 3, then $y=3$, in which case $z=5$, and $x=3$.
Therefore, the prime solution to the equation is
$$
x=3, \quad y=3, \quad z=5 .
$$