2. In the set of integers, solve the equation

$$
3^{2 a+1} b^{2}+1=2^{c}
$$
Solution. Let's first consider the case $a \geq 0$. Clearly $c \geq 0$ and $c=0$ implies $b=0$. We get that the solutions are $(a, 0,0)$ for any non-negative integer $a$. From the equation

$$
3^{2 a+1} b^{2}+1=2^{c}
$$

it follows that $b$ is an odd integer. The left side can be written in the following form

$$
3^{2 a+1} b^{2}+1=\left(3^{2 a+1}+1\right) b^{2}-(b-1)(b+1)
$$

For the right side of the last expression, note that $(b-1)(b+1)$ is divisible by 8, while $\left(3^{2 a+1}+1\right) b^{2}$ is divisible by 4 but not by 8. Therefore, $2^{c}=4$, i.e., $c=2$. Then, $3^{2 a+1} b^{2}=3$, so $a=0$ and $b= \pm 1$.

Now let's consider the case $a<0$. It is sufficient to consider the case $b>0$. Setting $d=-a$, the Diophantine equation from the problem condition takes the form

$$
\left(2^{c}-1\right) 3^{2 d-1}=b^{2}
$$

where $b, c$ and $d$ are natural numbers. Thus, $b$ is divisible by 3, which implies that $c$ is an even number. So $b=3^{d} x, c=2 y$, for some natural numbers $x$ and $y$. The Diophantine equation reduces to the form

$$
4^{y-1}+4^{y-2}+\ldots+1=x^{2}
$$

This implies $x=y=1$. Specifically, for $y \geq 2$ we would get $x^{2} \equiv 5(\bmod 8)$, which is impossible. Thus, in this case, the only solutions are $\left(a, 3^{-a}, 2\right)$, where $a$ is any negative integer.

The set $M$ of all solutions to the Diophantine equation from the problem condition can be described as follows:

$$
M=\{(a, 0,0) \mid a \in \mathbb{Z}\} \cup\left\{\left(a, \pm 3^{-a}, 2\right) \mid a \in \mathbb{Z}^{-} \cup\{0\}\right\}
$$