2.1. (New York, 77). Solve the equation $2^{x}+1=y^{2}$ in natural numbers.
2.1. Let's rewrite the equation as

$$
2^{x}=(y-1)(y+1)
$$

then for the sought values $x, y \in \mathbf{N}$, we have that the numbers $y-1, y+1 \in \mathbf{Z}^{+}$ are divisors of the number $2^{x}$, i.e., $y-1=2 p$ and $y+1=2^{q}$, where $p, q \in \mathbf{Z}^{+}, p<q$. Since the difference between $y-1$ and $y+1$ is 2, and the only positive divisor of 2 is 2 itself, we have $q-p=1$ and $2 p(2^{q}-2^{p})=2$, i.e., $p=1$, $q=2$. Finally, checking shows that the only possible value $y=3$ satisfies the equation only when $x=3$.