2. Determine all natural numbers $a$ and $b$ for which the number

$$
a^{4} b+3 b-2 a^{2} b^{2}-a^{2}-3 b^{3}
$$

is a power of two.
2. We have

$$
a^{4} b+3 b-2 a^{2} b^{2}-a^{2}-3 b^{3}=\left(a^{2}-3 b\right)\left(a^{2} b+b^{2}-1\right)
$$

the numbers $a^{2}-3 b$ and $a^{2} b+b^{2}-1$ must be powers of two. Clearly, $a=b=1$ is not a solution, so $a^{2} b+b^{2}-1>1$, and thus the number $a^{2} b+b^{2}-1$ must be even. This implies that $b$ is odd and $a$ is even, and then $a^{2}-3 b$ is also an odd number, i.e., $a^{2}-3 b=1$. Therefore, for some $k \in \mathbb{N}$, we have

$$
2^{k}=a^{2} b+b^{2}-1=a^{2} \frac{a^{2}-1}{3}+\left(\frac{a^{2}-1}{3}\right)^{2}-1=\frac{4 a^{4}-5 a^{2}-8}{9}
$$

From this, the number $5 a^{2}$ is even, so $a=2 t$ and $16 t^{4}-5 t^{2}-2=9 \cdot 2^{k-2}$ (in particular, $k \geqslant 2$). For $k=2$, we get $t^{2}\left(16 t^{2}-5\right)=11$, so $t=1$ (otherwise, the left side is greater than 11), and thus $a=2$ and $b=1$, which is a solution. Suppose further that $k \geqslant 3$. Then $5 t^{2}$ is divisible by 2, and thus by 4, so we have $9 \cdot 2^{k-2}=16 t^{4}-5 t^{2}-2 \equiv 2(\bmod 4)$, which implies $k=3$, i.e., $20=16 t^{4}-5 t^{2}=t^{2}\left(16 t^{2}-5\right)$. However, $t \geqslant 2$ (since $t$ is even), so the right side of the previous equation is at least 244; therefore, there is no solution in this case.

Thus, the only solution is $(a, b)=(2,1)$.