If $2^t = a^b \pm 1$, what are all the possible values of $t$? Let $t$ be a positive integer such that $2^t = a^b \pm 1$ for some integers $a$ and $b$, each greater than $1$. What are all the possible values of $t$?
The question is taken from here. I know the answer is $3$, so $a = 3$ and $b = 2$. But how can I prove that it's unique? Can you give me a hint, please? Thanks!
 A: The answer is $t=3$.
This answer proves that the following two claims are true :
Claim 1 :  If $t,a,b$ are integers such that $t\ge 1,a\ge 2$ and $b\ge 2$, then $2^t=a^b+1$ has no solutions.
Claim 2 :  If $t,a,b$ are integers such that $t\ge 1,a\ge 2$ and $b\ge 2$, then $2^t=a^b-1$ has the only one solution $(t,a,b)=(3,3,2)$.
Proof for Claim 1 : $a$ has to be odd, and $a^b\equiv -1\pmod 4$ implies that $a\equiv -1\pmod 4$ and that $b$ has to be odd. Since $b\ge 3$, we have
$$2^t=(a+1)(a^{b-1}-a^{b-2}+a^{b-3}-\cdots +1)\tag1$$
Here, note that 
$$a^{b-1}-a^{b-2}+a^{b-3}-\cdots +1=\frac{a^b+1}{a+1}\ge\frac{a^2+1}{a+1}\gt 1$$
and that
$$a^{b-1}-a^{b-2}+a^{b-3}-\cdots +1\equiv 1\pmod 2$$
It follows from $(1)$ that $2^t$ has an odd divisor greater than $1$, which is impossible. $\quad\square$
Proof for Claim 2 : $a$ has to be odd. 
If $b$ is even ($b=2m$), then $2^t=(a^m-1)(a^m+1)$. Since $a^m-1\ge 1$, there are non-negative integers $u,v\ (u\lt v)$ such that $a^m-1=2^u$ and $a^m+1=2^v$ from which $2=2^v-2^u$ follows. If $u=0$, then RHS is odd, which is impossible. If $u\ge 2$, then RHS is divisible by $4$, which is impossible. So, $u=1$ gives $a=3,m=1,v=2$.
If $b$ is odd, then since $b\ge 3$, we have $2^t=(a-1)(a^{b-1}+a^{b-2}+\cdots +1)$. We see that $a^{b-1}+a^{b-2}+\cdots +1$ is odd greater than $1$. It follows that $2^t$ has an odd divisor greater than $1$, which is impossible. $\quad\square$
