Test $\mathrm{C}$ Find the natural number solutions to the equation $2^{x}+1=y^{2}$.
(1977 New York Mathematics Contest)
(1977 American New York Mathematics Competition)
Solving the equation rewritten as $2^{x}=(y-1)(y+1)$, then for $x, y \in \mathbf{N}, y-1, y+1 \in \mathbf{Z}^{+}$ are factors of $2^{x}$, i.e., $y-1=2^{p}, y+1=2^{q}$, where $p, q \in \mathbf{Z}^{+}, p < q$. If $p \geq 1$, then $2^{p}$ will be a factor of 2, which is impossible, so $q=p+1$, and $2^{p}(2-1)=2$, i.e., $p=1, q=2$, thus $y=3$. Upon verification, $y=3, x=3$ is a solution to the equation.