3. Given $t \in \mathbf{N}_{+}$. If $2^{t}$ can be expressed as $a^{b} \pm 1$ (where $a$ and $b$ are integers greater than 1), please find all possible values of $t$ that satisfy the above condition.
3. Let $t$ be a positive integer such that $2^{t}=a^{b} \pm 1$. Clearly, $a$ is odd.
(1) If $b$ is odd, then
$$
2^{t}=(a \pm 1)\left(a^{b-1} \mp a^{b-2}+a^{b-3} \mp \cdots \mp a+1\right) \text {. }
$$

Since both $a$ and $b$ are odd, the sum or difference of an odd number of odd numbers is always odd, so
$$
a^{b-1} \mp a^{b-2}+a^{b-3} \mp \cdots \mp a+1
$$

is also odd. Therefore, it must be that
$$
a^{b-1} \mp a^{b-2}+a^{b-3} \mp \cdots \mp a+1=1 \text {, }
$$

which implies $2^{t}=a^{b} \pm 1=a \pm 1$.
Thus, $b=1$, which contradicts $b \geqslant 2$.
(2) If $b$ is even, let $b=2 m$. Then
$$
a^{b} \equiv 1(\bmod 4) \text {. }
$$

If $2^{t}=a^{b}+1$, then $2^{t}=a^{b}+1 \equiv 2(\bmod 4)$.
Thus, $t=1$. Hence, $a^{b}=2^{1}-1=1$, which is a contradiction.
If $2^{t}=a^{b}-1=\left(a^{m}-1\right)\left(a^{m}+1\right)$, the product of two consecutive even numbers can only be a power of 2 if
$$
a^{m}-1=2, a^{m}+1=4 \text {. }
$$

Thus, $a=3, b=2 m=2$.
Therefore, $2^{1}=a^{b}-1=3^{2}-1=8$.
In summary, the positive integer power of 2 that satisfies the given condition is $2^{3}$, i.e., $t=3$.