The expression you've coined reflects the orbit of one initial number $s$ towards $1$ by (the Syracuse-notation of) the Collatz-transformation. A perhaps better expression for this is
$$ a_{N+1} = \small {3^N a_1 + (3^{N-1} + 3^{N-2} \cdot 2^{A_1} + 3^{N-3} \cdot  2^{A_1+A_2} + ... + 3 \cdot  2^{A_1+A_2+...+A_{N-2} } + 2^{A_1+A_2+...+A_{N-2}+A_{N-1} } )\over 2^{A_1+A_2+...+A_{N-2}+A_{N-1} + A_N } }
$$
where the denominator is your expression $2^m$ .     
Let's make this monster-expression shorter; express the parenthese by the shortform $$ Q([A_1,A_2,...,A_N])=3^{N-1}+3^{N-2}\cdot 2^{A_1} + \cdots + 2^{A_1 +...+ A_{N-1}} $$ and $S = \sum_{k=1}^N A_k$ (I like the capital letters $N$ and $S$ and $A_k>0$ for the terms in exponents instead of small letters like $m$ as you use it here).        

Then we have in general 

$$ a_{N+1} = a_1 \cdot {3^N \over 2^S} + {Q([A_1,...,A_N])\over 2^S} $$

Your first question is to prove, that for all odd $a_1$ there is a vector $E(a_1)=[A_1,A_2,...,A_N]$ such that $a_1 \mapsto 1$ by finitely many $N$ steps.          
As it is well known, nobody has a proof so far and thus the Collatz-problem remains an `open problem` until now.                          

Your other observation is that of properties of three-step orbits ending at $1$. For this I propose to revert the notation: which numbers $a_3$ can be reached by the inverse Collatz-transformation of $N=3$ steps.              
We can write it this way:
$$ a_{k-1} = {a_k 2^{A_k}-1\over 3} \qquad \text{where } a_k \equiv 2^{-A_k} \pmod 3$$
and rewriting
$$ a_{N+1} = a_1 \cdot {3^N \over 2^S} + {Q([A_1,...,A_N])\over 2^S} \\\
 a_4 = a_1 \cdot {3^3 \over 2^S} + {Q([A_1,A_2,A_3])\over 2^S} \\\
 1 = a_1 \cdot {3^3 \over 2^S} + {Q([A_1,A_2,A_3])\over 2^S} \\\
 2^S = a_1 \cdot {3^3 } + {Q([A_1,A_2,A_3])} \\\
 2^S - {Q([A_1,A_2,A_3])} = a_1 \cdot {3^3 }  \\\
 {2^S - Q([A_1,A_2,A_3]) \over 3^3} = a_1   \\\
$$
Of course, ${1\cdot 2^{A_3}-1\over 3}$ being integer means $A_3=2k_3$ is even, and given your demand that $a_4=1$ gives $a_3={4^{k_3}-1\over3} = \{1,5,21,85,...\}$        
Along that line the possible values for $a_2$ and then for $a_1$ can be determined.   

In an older webpage I've drawn a tree where you can identify the possible $a_1$ starting at $a_4=1$ applying $3$-times the reverse odd steps . The vector of $[A_1,A_2,A_3]$ here is surely identical to what you have found yourself, but, well, there's not much to prove here: just to determine the possible values due to simple modular conditions (such that for instance $A_3$ must be even).