Let $f(x) = \dfrac{3x + 1}{2^{\nu_2(3x + 1)}}$. Clearly the $(1,2,4)$ cycle from the original Collatz map disappears under this context. So the Collatz conjecture-equivalent goal with $f$ is that there exists no tuples $(x_0, x_1, x_2, \dots, x_n)$ of odd natural numbers such that
$$\dfrac{3x_0 + 1}{2^{k_1}} = x_1, \ \dfrac{3x_1 + 1}{2^{k_2}} = x_2, \ \dots \ , \dfrac{3x_n + 1}{2^{k_0}} = x_0 \tag{1}$$
Notice that the indices wrap around $\pmod n$.
for some $k_i \geq 1$ and $n \geq 1$. We call such $(x,y)$ a $2$-cycle, and such $(x,y,z,w)$ a $4$-cycle for example, but we also call $(x,y,x,y)$ a $4$-cycle. Thus $2$-cycles "embed into" $4$-cycles. Clearly if there are no $k$-cycles (under this definition) then there are no $k'$ cycles for any $1 \lt k' \mid k$. Anyway, what I wanted to prove was the $2$-cycle Collatz special case.
Suppose that such $x_i, k_i$ existed. The sequence (1) defines a $2\times 2$ linear system (for the $2$-cycle case $n = 1$):
$$ -3x_0 - 1 + 2^{k_1} x_1 = 0 \\ 2^{k_0}x_0 - 3x_1 - 1 = 0 $$
The solution to this linear system is:
$$ \left\{\left( \frac{2^{k_{1}} + 3}{2^{k_{0}} \cdot 2^{k_{1}} - 9}, \ \frac{2^{k_{0}} + 3}{2^{k_{0}} \cdot 2^{k_{1}} - 9}\right)\right\} $$
That is to say, for each $(k_0, k_1) \in \Bbb{N}^2$ such that the entries to the tuple above are integral, the tuple forms an odd loop solution $(x_0,x_1)$, or a $2$-cycle.
Our goal is then to prove that the only possible solution is $(1,1)$.
But, if $(2^{k_0 + k_1} - 9) \mid (2^{k_1} + 3)$ then we have in particular that:
$$ 1 \leq 2^{k_0 + k_1} - 9 \leq 2^{k_1} + 3 $$
Now by algebraic rearrangement, we have equivalently:
$$ 10 \leq 2^{k_0 + k_1} \\ 2^{k_1}(2^{k_0} - 1) \leq 12 $$
Then $k_0 + k_1 \geq 4$ by the first inequality and $2^{k_1} \leq 12$. Which means $k_1 = 1, 2, 3$ since $k_1 \geq 1$ by definition. Similarly $(2^{k_0} - 1) \leq 12$ so that $k_0 = 1,2,3$. This completely determines all possibilities for the tuple $(k_0, k_1)$ (that we need to check for a $2$-cycle).
So that the possibilities are:
$$ (1,3),(2,2),(3,1),(2,3),(3,2),(3,3). $$
We have:
$$ \dfrac{2^{k_1} + 3}{2^{k_0 + k_1} - 9} = \\ (1,3): \dfrac{2^{3} + 3}{2^4 - 9} = \dfrac{11}{7} \\ (2,2): \dfrac{2^{2} + 3}{2^4 - 9} = \dfrac{7}{7} = 1 \\ (3,1): \dfrac{2^{1} + 3}{2^4 - 9} = \dfrac{5}{7} \\ (2,3): \dfrac{2^{3} + 3}{2^5 - 9} = \dfrac{11}{23} \\ (3,2): \dfrac{2^{2} + 3}{2^5 - 9} = \dfrac{7}{23} \\ (3,3): \dfrac{2^{3} + 3}{2^6 - 9} = \dfrac{11}{55} \\ $$
Since these are first components $x$ of a (rational) loop solution (an odd, accelerated $2$-cycle i.e.) $(x,y)$, you must plug back in and compute $f(x)$ to get $y$.
The only integral (and positive; note the $1\leq \dots$ used in finding them) loop solution is $(1,1)$ and we're done.
There exist no other odd $2$-cycles for the accelerated Collatz map. Which is to say, for the original Collatz iterations, there are no (positive) cycles with purely $2$ odd numbers in them.
So my question is a soft question and it's why then, historically, do people say that it was very difficult to prove the $2$-cycle case for Collatz? And I can't find any information on that proof, so if you can provide a link for us, that would help out us amateur mathematicians.
I'd propose, you call your "2-cycle" a "2-step-cycle" to avoid misunterstandings: the term "2-cycle" is already in use in the Collatz-discussion and means another thing and has already been disproved (J. Simons, ca 2000).
So, if you agree, I'll take your question as one on the "2-step-cycle": $$ b = { 3a+1\over 2^A } \qquad \rightarrow a = { 3b+1\over 2^B } \qquad \rightarrow b = { 3a+1\over 2^A }\qquad \rightarrow \cdots \tag 1$$ Build the product of the lhs in the above equations as well of the rhs: $$ b \cdot a = \left({ 3a+1\over 2^A }\right)\cdot\left({ 3b+1\over 2^B }\right) \tag 2 $$ Rearrange $$ 2^S (=2^{A+B}) = \left(3+{ 1\over a }\right)\cdot\left(3+{1\over b }\right) \tag 3 $$ We see, that the rhs can only get values between $16$ (when $a=b=1$) and $9$ (when $a=b=\infty$). But there are no perfect powers of $2$ between $9$ and $16$ except that which makes the trivial cycle ($a=b=1$) occur.
Back to the question, why "2-cycle disproof is difficult". Initially, Ray Steiner in 1976 discussed what he christened the "circuit", and then the "circuit which is also a cycle", whose existence he could then disprove (using "heavy tools" from transcendent number theory (Lagarias)).
This case has been re-christened "1-cycle" by John Simons in ca 2000 to lay ground for generalizations to "2-cycle", "3-cycle",..."k-cycle".
For me, the nicest method to get intuition about this, is my notation $$ T(a,[A]) := a_2= {3a+1\over 2^A} \\ T(a_1,[A_1,A_2]) :=a_3= {3{3a_1+1\over 2^{A_1}}+1 \over 2^{A_2}} \tag 4 $$ and so on, generalizable to as many exponents $A_k$ as wanted.
With this, Steiner's "circuit which cycles" (and Simon's "1-cycle") is defined by $$ a_1 = T(a_1,[1,1,1,1,1,...,1,A]) \qquad A\gt 1 \tag 5$$ with as many "1" as you want (I say $N$ for the (N)umber of all exponents/steps and $S$ for their (S)um).
It has been much difficult to find an argument, that no such "1-cycle" exist except that with no exponents $1$ and only $A=2$: $ 1 = T(1,[2]) $ (and only in the negative numbers $-1 = T(-1,[1]) $ and $-5=T(-5,[1,2])$ ).
The "2-cycle" is now a construction $$ a_1 = T(a_1,[1,1,1,1,1,...,1,A,1,1,1....,1,B]) \qquad A,B\gt 1 \tag 6$$ with as many "1" you want in each sub-trajectory, and J. Simons needed a dozen-page paper to do the disproof in the analoguous style as R. Steiner had it invented.
Generalizing this, J. Simons ( & B.de Weger) could come up with disproofs of "k-cycles" up to $k=68$ (or $74$, I'm not up-to-date).
What is, btw., not so well known is that the third known cycle in the negative numbers beginning at $-17$ is a nice, true "2-cycle" having, if I don't err $-17=T(-17,[1,1,1,2,1,1,4])$ (and I like this small fact as a little gem in all this discussions :-) ).
update I forgot, I have a very old workout of this (and a bit more) when I've learned this things myself in 2004. Perhaps you find even more impulses for further thinking/fiddling in that pages/subpages. See subsections on "loops" on my homepage.