The cycle problem is simple enough for an amateurish approach. It is easy to find out, that a cycle must be longer than some minimal length. However, the only proof so far, which shows, that a cycle cannot occur even if arbitrary length is assumed, was done by means of transcendental number theory and the concept of rational approximation. Also, this could only be used for a special simple type of cycles. R. Steiner succeeded around [1977][3] with the first proof of this type (disproving the "1-cycle"); J.Simons 1996 and later in cooperation with B.de Weger (since 2002)could extend this concept up to "68-cycles", a 68-fold concatenation of arbitrary long partial sequences of the same type. After that the currently best known bounds for that approximation are too weak, and Simons/deWeger assume, that finally some completely different method must be found to proceed fundamentally here.
[update] Perhaps I should make the relation of your question to the method of Steiner/Simons (and my own) clearer here.
The mentioned Steiner and Simons/deWeger theorems concern cycles of the general form: (ID)(ID)(ID)...(ID) DD...DD with u occurences of ID and d occurences of D, so u upwards steps followed by d downwards steps. To make this simpler we would denote the Steiner description with one letter, say U, for the combination ID.
So the general form were U...U D...D which is called "1-cycle" if this describes a cycle.
Then first, Steiner showed 1977 there is no cycle how many (u and d) U's (or (ID)'s in your notation) and D's in that specific arrangement occur.
Now let's denote one of that U...U D...D-constructions by $C^{u,d}$ where u again denotes the number of U and d the number of D. John Simons showed then 2004 that no cycle of the twofold concatenated form U...U D...D U...U D...D = $C^{u_1,d_1}C^{u_2,d_2}$ exists (besides of the "trivial" cycle).
The relation of your notation to the 3-, 4-, and up to the 68-cycle is then the obvious, I think.
If you look at my own article you'll find, that I reduce, for instance I DDD I DD, similarly to $b=T(a;3,2)$ - just counting the number of D's after one I, and introducing a and b as formal start- and endvalues of the full sequence of transformations, and analyze this in terms of powers of 3 and 2 as exponential diophantine expressions. This leads to very interesting (and more general) properties of the relation of powers of 2 with powers of 3.
Appendix to avoid a frequent misunderstanding
What Steiner and Simons did call a "1-cycle" is not identical to the "1-step-cycle", but is of arbitrary length/steps, however with a nice analytically exploitable structure.
A disproof for the "1-step-cycle" is indeed as simple as Robert Frost in his comment mentions: Assume "1-step" as $b=(3a+1)/2^A$ and this being a cycle we must have $b=a$ and thus $a=(3a+1)/2^A$. This can be reformulated
$$ a = {3a+1\over 2^A} \to a2^A = 3a+1 \to a = {1\over 2^A-3}
$$
and indeed, there is only one solution for positive integer numbers $a$ possible, namely $A=2,a=1$ - giving the trivial cycle.
Moreover, this can be extended easily to disprove manually a small set of "$N$-step-cycles":
Let $N=2$ and $b=(3a+1)/2^A$, $c=(3b+1)/2^B$ and to make it a cycle, $c=a$. Then we can make the product
$$ a \cdot b = \left( {3a+ 1\over 2^A}\right) \left( {3b+ 1\over 2^B}\right)\\
\to 2^{A+B} = \left( 3+{1\over a}\right) \left( 3+{1 \over b}\right)
$$
Of course, the rhs can only be between $9$ and $16$, where if it is $16$ we must have $A+B=4$ and $a=b=1$ thus having the trivial cycle only.
Let $N=3$ and $b=(3a+1)/2^A$, $c=(3b+1)/2^B$ , $d=(3c+1)/2^C$ and to make it a cycle, $d=a$. Then we can make the product
$$ \to 2^{A+B+C} = \left( 3+{1\over a}\right) \left( 3+{1 \over b}\right)\left( 3+{1 \over c}\right)
$$
and now the rhs can only be between $27$ and $64$. Being equal to $64$ requires $a=b=c=1$ being the trivial cycle, which we are not interested in. Another solution to equal a perfect power of $2$ , namely $2^{A+B+C}$ it must equal $32$. But now, if $a,b,c \ne 1$ and all different and none divisible by $3$ we have at most
$$ \to 2^{A+B+C} = \left( 3+{1\over 5}\right) \left( 3+{1 \over 7}\right)\left( 3+{1 \over 11}\right)
$$
and we see, that this is smaller than the required $32$. But increasing $a,b,c$ doesn't help since then the rhs decreases even more, so no "3-step-cycle" is possible. (see some more examples in my small online-treatize)
In the same way a lot of attempted "N-step-cycles" with fixed, small and large $N$ can be disproved. However, the Steiner's and Simon's $1-cycle$ concept includes all arbitrary large $N$ - and to have a disproof in such a generality we need a more general method of disproof (and also the specific structure of the problem when we have a single "up--down" pattern in the $a_1,a_2,a_3,...,a_N$ elements of an assumed cycle only).
[3]:
http://???/ Steiner, R.P.; "A theorem on the syracuse problem", Proceedings of the 7th Manitoba Conference on Numerical Mathematics ,pages 553–559, 1977.
Let's note that this is not a question of whether Collatz is undecidable.
The statement $\neg\mathrm{Con}(PA)$ is undecidable (by $PA$, assuming $PA$ is consistent) but nevertheless $\neg\mathrm{Con}(PA)$ is provably equivalent to a certain Turing machine halting (the one that searches for a proof of a contradiction in PA).
Rather, the question is whether
there is a $\Pi^0_1$ statement $\varphi$ such that the Collatz problem, which on its face is $\Pi^0_2$, is already known to be equivalent to $\varphi$.
Here already known means in particular that we are not allowed to assume that Collatz is or is not provable or disprovable in any particular system, unless we already know that.
The best evidence that there is no such $\varphi$ seems to be in the paper mentioned by @Burak:
Kurtz, Stuart A.; Simon, Janos, The undecidability of the generalized Collatz problem, Cai, Jin-Yi (ed.) et al., Theory and applications of models of computation. 4th international conference, TAMC 2007, Shanghai, China, May 22--25, 2007. Proceedings. Berlin: Springer (ISBN 978-3-540-72503-9/pbk). Lecture Notes in Computer Science 4484, 542-553 (2007). ZBL1198.03043.
Namely, they give a parametrized family of similar problems such that the collection of parameters for which Collatz-for-those-parameters is true, is $\Pi^0_2$-complete and hence not $\Pi^0_1$.
They can do this without thereby solving the Collatz problem, just like Matiyasevich et al. could show that solvability of diophantine equation was $\Sigma^0_1$-complete, without thereby solving any particular equation themselves.
If Collatz could somehow be simplified to a $\Pi^0_1$ form then quite plausibly the generalized version could too by the same argument (whatever that hypothetical argument would be) but that Kurtz and Simon show will not happen.
Best Answer
[update 2] Rereading the question after my own answer I note, that the doubly used symbol m for length of a series of collatz-transformation and the m for the minimal element (M for the maximal) irritated me and made me misunderstanding the question.
Anyway - the already given answer might be interesting for the background of my argument below as well as for further analysis of the problem at hand, so I'll leave it here for the cursory reader.
To answer the real question : "is there some requirement, that in a cyclic Collatz-trajectory the maximal element is a perfect-power-of-2-multiple of the minimal element (M=2^k m) ?" : there is no such requirement to all my knowledge.
Also this would only define a single special case according to my notation for a cycle, which by convention might begin with the smallest element m : $\small m = T(m; A_1,A_2,\ldots ,A_{n-1},A_n) $.
Here your requirement simply means, that the maximal element M must occur after the transformation $\small A_{n-1} $ (whose exponent must thus equal 1) frome where then the $\small A_n$ is the k in M=2^k m in your question. I didn't come across such a restriction in my own study or in any seen literature, and cannot see any reason for it (see derivations below). [end update 2]
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 collatz-steps (m=N + S in my notation, where I refer -writing "steps"- to the number N of abbreviated steps $\small T(a;A)$
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 N=1, S=2 and m=N+S=3 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 N (and respectively 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) $ . Here m=N+S=2+4=6 .
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 Collatz steps" interspersed; even less can the distance between the minimal and maximal member of a loop be $\small 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 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 Collatz 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. The 2-step cycle was $\small 2^{A+B} = (3+1/a)(3+1/b) $ and allowing negative a and b gives the example: $\small 2^{A+B} = (3-1/5)(3-1/7) = {14 \over 5} \cdot {20 \over 7} = {8 \over 1} = 2^3 $ which is a perfect power of 2 as required. Then $\small A+B=3$ and one of them must equal 1. In fact we have the 2-step-cycle $\small -5=T(-5;1,2) $
Finally: 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