2. In the set of integers, solve the equation $x^{2}=3^{y}+7$.

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Solution. For $y=0$ and $y=1$, we have $x^{2}=8$ and $x^{2}=10$ which do not have solutions in the set of integers, so $y \in \mathbb{N} \backslash\{1\}$. Let's note that $3^{y}+7$ is an even number, so $x^{2}$ is even, and therefore $x$ is even. From this, it follows that the right side of the equation is divisible by 4. Let $y$ be odd, i.e., $y=2 k+1$.

Then $x^{2}=3^{2 k+1}+7=3 \cdot 3^{2 k}-3+10=3\left(3^{k}-1\right)\left(3^{k}+1\right)+10$. Let's note that $x^{2}$ and $3\left(3^{k}-1\right)\left(3^{k}+1\right)$ are divisible by 4, but 10 is not. Due to the obtained contradiction, we conclude that $y$ must be an even number. Let $y=2 s, s \geq 1$.

Now, $x^{2}=3^{2 s}+7$, or $\left(x-3^{s}\right)\left(x+3^{s}\right)=7$, where $x-3^{s}<x+3^{s}$. Then only the following two cases are possible:

$$
\left\{\begin{array}{l}
x-3^{s}=1 \\
x+3^{s}=7
\end{array}\right.
$$

from which $x=4, s=1$ and

$$
\left\{\begin{array}{l}
x-3^{s}=-7 \\
x+3^{s}=-1
\end{array}\right.
$$

from which $x=-4, s=1$.

Possible solutions to the initial equation are $(4,2),(-4,2)$, and upon verification, these are indeed solutions to the initial equation.