And here is an attempt to get closer to a general proof that any solution to the equation 2^k -3^x = m with 0<m< c x implies the existence of a non-trivial Collatz cycle. We leave the positive constant c \leq 1 undetermined for now …
Thus, we assume the existence of such m \leq cx. It is not difficult to see that m is odd and is not a multiple of 3.
The case k > \frac{\log 3}{\log 2} \, x + 1 is easy to rule out since it implies
2^k > 2^{\frac{\log 3}{\log 2} \, x + 1} = 2 \cdot 3^x > 3^x + c x.
For the case k = \left\lceil \frac{\log 3}{\log 2} \, x \right\rceil, we call v_0 the vector matching the first k digits of the upper Sturmian word of slope \frac{\log 2}{\log 3} starting as
(110)^3(10)(110)^2(10)…
The number of 1’s in v_0 is equal to x. Since there are no two consecutive 0’s in v_0, the number of occurrences of 10 is equal to the number z of 0’s:
z = k-x > \left( \frac{\log 3}{\log 2} - 1\right) x > 0.58 x.
Next, we define the vectors v_i obtained by flipping the i^{\rm th} occurence of 10 in v_0 for i=1,\ldots,z.
Express the differences d_i = \beta(v_i)-\beta(v_0) under the form
d_i = 2^{a_i} \cdot 3^{x-1-b_i}
where
1=a_1 < a_2 < \ldots < a_z < k
1=b_1 < b_2 < \ldots < b_z < x
Empirically, we also have a_i = \left\lfloor \frac{\log 3}{\log 2} \, b_i \right\rfloor.
We want to show that all non-zero congruence classes modulo m are covered by the sequence of d_i values, which is only possible if m < z . So we set c = \frac{\log 3}{\log 2} - 1 \simeq 0.58.
Observe that d_{i+1} - d_i = \pm 2^{a_i} \, 3^{x-1-b_{i+1}} is never divisible by m (unless m=1), so that d_i and d_{i+1} are not in the same class for any i.
More generally, for i<j, we have d_j - d_i = 2^{a_i} \, 3^{x-1-b_j} \left( 2^{a_j-a_i} - 3^{b_j-b_i} \right) which is divisible by m iff \delta_{i,j} = 2^{a_j-a_i} - 3^{b_j-b_i} is divisible by m. The question is now whether the numbers of the form \delta_{i,j} can be divisible by m = 2^k-3^x.
Here we can use the fact that whenever 2^k-3^x divides 2^A-3^B, then it also divides 2^{k-nA}-3^{x-nB} for any n as long as the exponents are non-negative. If we repeat this process, we should be able to rule out this case. However, it is far from straightforward to finalize this proof and I leave it as a challenge…