Example 19 (18th All-Russian Mathematical Olympiad) Find the prime solutions of the equation $2^{x+1}+y^{2}=z^{2}$.

Find the prime solutions of the equation $2^{x+1}+y^{2}=z^{2}$.
From the given equation, we have $2^{x+1}=(z+y)(z-y)(z>y)$.
Thus, we have $\left\{\begin{array}{l}z-y=2^{k}, \\ z+y=2^{x+1-k},\end{array}\right.$
where $k \in \mathbf{Z}$ and $0 \leqslant k \leqslant x$. Since $k-1 \geqslant 1$, $y$ and $z$ cannot both be prime numbers. If $k=1$, then
$$
y=2^{x-1}-1, z=2^{x-1}+1 \text {. }
$$

Since $2^{x-1}-1, 2^{x-1}, 2^{x-1}+1$ are three consecutive positive integers, one of them must be divisible by 3, but $2^{x-1}$ cannot be divisible by 3, so only $y$ or $z$ can be divisible by 3.
If $y$ is divisible by 3, then $y=3, x=3, z=5$.
If $z$ is divisible by 3, then $z=3, x=2, y=1$ is not a prime number.
Therefore, the prime solution to the equation is $x=3, y=3, z=5$.
(2) Congruence Method (including parity analysis)
The congruence method involves taking a special modulus on both sides of the equation to establish a congruence (specifically, when the modulus is 2, it is parity analysis), narrowing the range of variable values, thereby deriving the integer solutions of the indeterminate equation or leading to a contradiction to determine that it has no solution.