If you write a collatz-transformation this way $ \small a_{k+1} = (3 a_k + 1)/2^A $ where A is the number of all following even (divide-by-2) steps, then also all $\small a_k$ must be odd.
With this write for one transformation $\small a_{k+1} = T(a_k;A) $ and for more $\small a_{k+m} = T(a_k;A_1,A_2,\ldots,A_m) $.
For $\small A=1 $ these steps are increasing and if $\small A>1$ the steps are decreasing.
A cycle occurs, if $\small a_m=T(a_0;A_0,A_1,\ldots,A_m) = a_0 $ thus $\small a_m = a_0 $.
Now let's look at two different types of cycles - the "general" one, where the exponents $\small A_k $ have no a priori restriction, and the "primitive" one, where all but the last $ \small A_k = 1 $, thus $\small a_0 = T(a_0;1,1,1,1,\ldots,1,A) $ with say $\small N-1 $ times 1 and only one $\small A_{N} \gt 1$
For the primitive cycle of the length N the smallest element $\small a_0 $ must have the form $\small a_0 = 2^{N-1} \cdot 2w - 1$ where w is some positive odd integer and that primitive cycle increases then up to $\small a_{N-1} = 3^{N-1}\cdot 2w-1 $ from where it must decrease by a consecutive set of A "even" transformations.
The proof of Ray Steiner (and the subsequent proofs of J.Simons and B.de Weger) use that requirement of the "primitive cycle" (in their nomenclature "1-cycle", see wikipedia) to show that such a cycle cannot exist except $\small a_0 = T(a_0;2) \qquad a_0=1 $ (where no exponents of value 1 occur - the degenerate case (also called "circuit").
For the general cycle this is much more complicated and does definitely not have the properties which you sketch in your op.
[added] Answering to the comment. A characterization of the general cycle of m steps
Assume again one step as on the form $\small a_{k+1} = (3 a_k +1)/2^{ A_k} = T(a_k;A_k) $. Then to have a cycle this means (ignore the index k here) $\small a=(3a +1)/2^A $ and also $\small 2^A=(3a +1)/a = (3 + 1/a) $ This can only be solved if a=1 and A=2, so there is only one general-cycle of length m=1 and we see a characteristic of the exponent A, which is much descriptive: the number of even steps of the original notation of the collatz-transformation is 2.
Now we assume a 2-step cycle $\small a = T(a;A,B) $ and dissolve this in two steps: $\small b=T(a;A) \qquad a=T(b;B) $ thus $\small b = (3 a +1)/2^A \qquad a=(3b+1)/2^B $. I'm used to write S for the sum A+B meaning the whole number of even steps, and N for the number of "odd steps", both wrt the original Collatz-notation, and in my notation N is the number of steps (and the power of 3 involved). If we write the (trivial) product of the two involved elements a and b in their direct notation and in their transformed expression we get $\small a\cdot b = (3b+1)/2^A \cdot (3a+1)/2^B $ and this can be rewritten as $\small 2^{A+B} = (3+1/a)(3+1/b) $ This is an interesting form and easily generalized for the analysis with bigger m . Here we see, that a 2-step general cycle can only exist if $\small (3+1/a)(3+1/b) $ is a perfect power of 2; now considering the whole set of odd positive integers for a and b we see, that that product can vary only between $\small 9 \ldots 16 $ and thus must be S=A+B=4 and this requires a = b = 1 and thus A=B=2 (which is then only a concatenation of the trivial cycle $\small 1=T(1;2,2) $ .
This way you may proceed studying longer assumed cycles. For instance it shows that the whole distance between two numbers $\small a_k - a_j = 2^S$ where S is the number of even transformations can never occur because there are always "odd steps" interspersed; even less can the distance between the minimal and maximal member of a loop be $\small N+S 2^{N+S} = 2^m $, which were the whole length of the cycle in the counting of original Collatz-steps.
This formula describes pretty well important properties of the exponents of 2 in relation to the length of a "general cycle" , so it might be useful Howerver for the answering of your question. However - I can't relate anything in that formula to the asked property of a connection between the distance minimal...maximal member and 2^m where m is the number of all steps and actually is $\small m=S+N$ nd so I think there is none. (
Addendum : It might be interesting to look at the existing cycles in the domain of negative odd numbers; they can also be identified using the procedere exercised above. (Also a short treatize using the notation here is in that article of mine)
1 Steiner,R.P.; A theorem on the syracuse problem, Proceedings of the 7th Manitoba Conference on Numerical Mathematics,pages 553...559, 1977
