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
Dan's algebraic justification is correct, but you may get more intuition about why this is happening from the above picture. Each time you want to enlarge the square by one unit, you have to add an extra row, an extra column, and one more square to fill in the corner. These correspond directly to the $n+n+1=2n+1$ that Dan mentioned. And of course, $2n+1$ is how odd numbers look.
Looking at it from the other direction, you can use the same idea to convince yourself of Noah's claim that $1+3+5+\dots+(2n-1) = n^2$. Imagine the left-hand side as representing the upper sequence of green and orange squares. You add on first 1 tile, then 3, then 5, and so on to construct larger and larger squares. At each step, you are adding one row (n) and one column (another n), then removing that one tile in the top right corner where they overlap (for a total of 2n-1).