The so-called "1-cycle" in the Collatz-problem was already disproved by Ray Steiner 1977. However, he used transcendental number theory to achieve that, and Lagarias commented, it is surprising that such heavy weapons are needed for such a tiny result.
I think I have an elementary proof now but usually in the Collatz-problem your proofs are faulty, so I better ask for a crosscheck. As proposed in the MSE.meta I'll put the problem-description in this question and post my attempted proof in an own answer.
The notion of 1-cycles begin with the definition of increasing steps $a_{k+1} = {3a_k +1\over 2} $ if $a_k$ is odd and decreasing steps $a_{k+1} = {a_k\over 2} $ if $a_k$ is even.
If we consider the simple type of a cycle, which occurs if $N$ increasing steps are followed by $B$ decreasing steps, then let us call this with Steiner, Simons, de Weger and others the "1-cycle" of length $S=N+B$.
Notes: 1) here the $N$ contains the number of increasing steps and defines, that the term $3^N$ will become involved. It is my main varying parameter.
2) Depending on $N$, there is $S$ such that $2^S \gt 3^N \gt 2^{S-1}$ so in fact $S$ is a function of $N$, and can also be determined by $S(N)=\lceil N \log(3)/\log(2) \rceil$ . I used the symbol $S$ here to indicate it is the sum of the exponents of all powers of $2$ which are involved and also the shorter form $S$ instead $S(N)$ .
3) I introduce also the short-name for the difference $B = S-N$, thus also depending on $N$ . $B$ is simply the number of purely decreasing steps after the $N$ purely increasing steps.
With Steiner, we can state, that an initial number a which we wish to increase by N steps must have the form $a = 2^N \cdot k -1$ ; here we can use any odd integer $k \gt 0$ . The result of that transformation, let's call it b has then the form $b=3^N \cdot k -1 $ with the same parameter k.
If b is then transformed by B decreasing steps and reaches c then c has the form $c={3^N \cdot k\ - 1 \over 2^B}$
If we want a cycle then this requires $c=a$ and this is also the "1-cycle" where $a$ is the minimal element.
Well, thus identifying a with c this gives the "critical formula" $ \tag{1.0} 2^N\cdot k - 1 = {3^N \cdot k -1 \over 2^B}$ Here we assume some $N$, get from this the dependend variable $B$ and from this try to find some or one admissible $k$.
Simply rearranging gives the more focused form which appears (using different letters) in J.Simon's paraphrase of R.Steiner's proof: $ \tag{1.1} (2^S - 3^N)\cdot k = 2^B - 1 $ (remember $S=N+B$ and $2^S \gt 3^N$ is understood). Then the relevant statement is:
For a 1-cycle of any length $N$ to exist we must have a solution in positive integers $(N,S,B,k)$ in eq. (1.1) .
$\qquad \qquad$ (two other useful representations of that critical equation are $ \small (2^S - 3^N)\cdot k = (2^S - 2^N )/2^N \quad , 2^S \gt 3^N \quad \text{ or }\\ \small 2^B = {2^S\over 2^N}={3^N \cdot k -1 \over 2^N\cdot k-1}$
Using results of transcendental number theory and rational approximation with linear forms of logarithms Steiner proved, that there is no $2^B$ satisfying that formula except if $N=1$ and from this $S=2$,$B=1$,$k=1$ and $ a=1 \Rightarrow b = 2 = {3a+1\over 2} \Rightarrow a=1={b \over 2} $ which is called the "trivial cycle".
Question: how can it be proven by elementary means, that the formula (1.1) has not other solutions?
Actually, this question is asking for help: to crosscheck my proof attempt, which I'll state in an answer below.
The Steiner-proof is not online, but John Simons has used it when he extended the Steiner-method for the 2-cycle. His article which contains a paraphrase is online at AMS . Note that I use different letters for the variables to be consistent with my own older discussions
[Added]: Empirically, the last shown form of the critical equation can never be satisfied for $N \gt 1, k \ge 1$ because the rhs falls empirically between the sharp bounds $ \small {3^N\over 2^N} \lt {3^N \cdot k -1 \over 2^N\cdot k-1} \lt \Big\lceil {3^N \over 2^N} \Big\rceil \qquad \text{ for }K=1 \ldots \infty$
(and thus is never arriving the next integer above $(3/2)^N$) where the right inequality conforms with that conjectured bound in the Waring-problem, but which is not yet proven.