I propose a proof of k = \lceil x \log_2(3) \rceil for a larger set of potential cycles. The only conditions are that x \geq 27 and \frac{M}{m} \leq 7 with m and M the least and greatest odd elements of the cycle, respectively. In other words, the cycle has to be relatively flat, unlike a circuit, and not too short.
Amazingly, the main argument is that there is simply not enough space between m and M for a cycle of length k \neq \lceil x \log_2(3) \rceil with x odd terms.
The proof is elementary 
and goes as follows:
Let a = \frac{M}{m} > 1 .
We assume that
- a \leq 7;
- k = \lceil x \log_2(3) \rceil + d with d \geq 0 and x \geq 27.
Since the cycle has x odd terms coprime to 3 ranging between m and M, we estimate the number of such integers in this range which is at most \frac{M-m}{3} + 1 = \frac{a-1}{3}m + 1. We deduce that x \leq \frac{a-1}{3}m + 1, which we can also write
m \geq \frac{3(x-1)}{a-1}.
Next, we consider the well-known equality
2^k = \prod_{i=1}^x (3+\frac{1}{n_i}) \quad
holding for all cycles with odd terms n_1, \ldots, n_x. See Lemma 2.2 in Eliahou’s paper for an elementary proof.
It implies that: 2^k \leq \left(3+\frac{1}{m}\right)^x.
On the other hand, we have 2^k > 3^x \, 2^d.
It yields
2^d < \left(1 + \frac{1}{3m}\right)^x .
Then, we take the logarithm:
d \log 2 < x \log \left(1 + \frac{1}{3m} \right) < \frac{x}{3m} < \frac{(a-1)x}{9(x-1)}
so that
a > 1 + \frac{9(x-1)d\log 2}{x} \geq 1 + \frac{9 \cdot26 \, d\log 2}{27} > 1 + 6 d.
Our very first assumption that a \leq 7 implies d=0.
\square
Consequently, we can answer in the two extreme cases of a circuit and a high cycle (both of which we know do not exist), but also in the case of all sufficiently “flat” (high?) cycles. If anyone was wondering!