I've not enough time to complete that answer at the moment, but I can leave you in a position where you likely can proceed on your own.
The factoring-out is a good start: You get
$$ 2^b(2^A-1) = 3^d(3^C-1) $$
Then you rewrite
$$ {2^A-1\over 3^d} = {3^C-1 \over 2^b} $$
[lhs]: Then there is a little rule (see "Euler's totient theorem") , which values $A$ must assume for the numerator in the lhs to be divisible by $3$ or more precisely exactly by $3^d$:
$$A = \varphi(3^d) = 2 \cdot 3^{d-1} \qquad \text{ by Euler's totient theorem}$$
Sidenote: in this case, we deal with the primefactor $3$ and the $\varphi(3^d)$ is indeed the "order of the cyclic subgroup of $2^n \pmod 3$ - which is relevant here, since we want the exponent $A$ such that the denominator divides exactly. The situation here, with primefactor $3$ is easier than (and different from) some other primefactors like for instance $p=7$ where the $\varphi(7)=6$ but the order of the cyclic subgroup $ \pmod 7$ is smaller and actually $\lambda_7=3$ (Note, this is not the Carmichael-function!) .
One can multiply additional factors to $A$ which only must not contain the primefactor 3: with $v$ where $\gcd(v,3)=1$ the complete expression is
$$A = \varphi(3^d) \cdot v = 2 \cdot 3^{d-1} \cdot v \qquad \qquad \gcd(v,3)=1 \tag 1$$
[rhs]: Similarly this is for the rhs:
$$ {C = 1 \cdot w \text{ for } b=1 \text{ and } \qquad \qquad \gcd(w,2)=0 \tag {2.1} } $$ $$ C=2^{b-2} \cdot w \text{ for } b \ge 3 \qquad \qquad \gcd(w,2)=0 \tag {2.2}$$
Here a solution for $b=2$ does not exist; because any $3^C-1$ which is divisible by $2^2$ is also divisible by $2^3$ and thus one factor $2$ remains, and then this cannot equal the lhs which has an odd value. In effect, this blows up our equation to
$$ {2^{2 \cdot 3^{d-1} \cdot v} - 1\over 3^d} = {3^{1 \cdot w}-1 \over 2^1} \tag {3.1 when $b=1$}$$
$$ {2^{2 \cdot 3^{d-1} \cdot v} - 1\over 3^d} = {3^{2^{b-2} \cdot w}-1 \over 2^b} \tag {3.2 when $b\ge 3$ }$$
I'm not going further at the moment; but now you might look for existence of different additionally primefactors on the lhs and the rhs for choices of $d$ and $b$ and $v$ and $w$. For instance, if $d \gt 1$ we have in the lhs additional primefactors with group-order $3,6,9,18$ which means the primefactors $7,-,73,19$ which must then also occur in the rhs. We know by a theorem of Szigmondy(?spell) that only for $A=2\cdot 3=6$ there is no other additional ("primitive") primefactor existent. This and the "trivial" cases should come out as the only possible solutions.
[update]
To make Jack's observation more explicite. First we observe, that in the exponent of the LHS there is unconditionally a factor 2 , so to improve readability a bit we write
$$ {4^{ 3^{d-1} \cdot v} - 1\over 3^d} = {3^{2^{b-2} \cdot w}-1 \over 2^b} \tag {3.2}$$
were we took the second version, assuming $b \ge 3$
Then we observe the table of cycle-lengthes for the first few primes for the numerator in the LHS and the RHS:
pf 4^A-1 3^C-1
-------------------
2 - 1
3 1 -
5 2 4
7 3 6
11 5 5
13 6 3
17 4 16
19 9 18
23 11 11
29 14 28
31 5 30
37 18 18
41 10 8
43 7 42
47 23 23
53 26 52
59 29 29
61 30 10
67 33 22
71 35 35
73 9 12
... ... ...
Now we discuss the possibility of equality, given that $d=2$. Then we look at the composition of $A$, see, what primefactors on the lhs are involved, conclude that they are involved in the rhs (note: must be to the same power!), that sets conditions on $C$ which includes more primefactors for the rhs thus also for the lhs, which then imposes a new composition of A, and so on, until we possibly get a contradiction. Here is a short decision-path-diagram. The "*" sign marks the inclusion of the primefactor due to compositions of $A$ or $C$:
A=3^1
|
* 3 1 0
* 7 3 6 ==> C=2.3.w
|
* 13 6 3 ==> A=2.3
|
* 5 2 4 ==> C=4.3.w
|
* 73 9 12 ==> A=2.3^2 #####
contradiction, because exponent of 3 in A was assumed=1
So for the assumption of $d=2$ there is no solution of the original problem
Best Answer
There are infinitely many counterexamples.
Consider $n\pmod {24}$. We consider a covering system for such residues. Specifically, remark that:
$$n\equiv 1 \pmod 2\implies 2^n\equiv 2\pmod 3$$ $$n\equiv 2 \pmod 4\implies 2^n\equiv 4\pmod 5$$ $$n\equiv 1 \pmod 3\implies 2^n\equiv 2\pmod 7$$ $$n\equiv 4 \pmod 8\implies 2^n\equiv 16\pmod {17}$$ $$n\equiv 8 \pmod {12}\implies 2^n\equiv 9\pmod {13}$$ $$n\equiv 0 \pmod {24}\implies 2^n\equiv 1\pmod {241}$$
These are easily verified, case by case. And it is easily verified that each $n\in \mathbb N$ satisfies at least one of the congruences.
Thus we are lead to consider: $$\text{ChineseRemainder[\{1,1,5,1,4,240\},\{3,5,7,17,13,241\}]}=1518781$$
Any prime congruent to that $\pmod {3\times 5\times 7\times 13\times 17\times 241}$ will be a counterexample, and by Dirichlet there are infinitely many of these.
The least such prime is $68627641$, followed by $202845361$.
Of course, there might be smaller counterexamples...but these counterexamples are guaranteed to allow small prime factors for every $n$.