I am kind of new to this problem and I tried solving it with open mind. Please don't be judgmental, this is what I got.
Let us assume, for the sake of contradiction, that the Collatz conjecture is false and that there exists a cycle in the Collatz sequence other than the well-known $4-2-1$ loop. Consider this hypothetical cycle, which must have a minimum and a maximum points.
We focus our attention solely on the orbit from the minimum point to the maximum point. We define a local minimum point as a point that once we reach it in the orbit the value cannot decrease any further as we continue along the orbit. It is important to note three facts about these local minimum points:
1.In Collatz sequence generated between any two numbers f and g, f<g there exist at least one local minimum point. If there is no such local minimum point then f itself serves as local minimum.
Local minimum points must be odd, as even numbers would decrease upon division by $2$, contradicting the definition of a local minimum.
Between two consecutive local minimum points, there can be no numbers in their sequence that connect them, as any such point would also create a local minimum (*according to 1) , thus violating its definition.
Let us consider a local minimum point $r$. Since $r$ is odd, multiplying it by $3$ and adding $1$ yields an even number. Dividing this even number by $2$ gives us $1.5r+0.5$. Therefore, the next local minimum point must lie somewhere in the range between $r+1$ and $1.5r+0.5$, inclusive. If no such point exists in this range, it contradicts our assumption, as it would indicate a decrease below the minimum of the local minimum.
The only exception to this rule is the last local minimum point before reaching the maximum point. For simplification let us denote the maximum point as $9p+1$, and the last local minimum point as $3p$ (were $p$ belongs to Real as no numbers in the sequence divide by 3), which has an orbit length of $1$ and rises.
Now, let us examine the second highest local minimum point in the orbit, denoted as $2p+a$, where $a$ is an integer lower than $p$. Following the path from this local minimum, applying the transformations of $3n+1$ and division by $2$, we arrive at the expression $3p +1.5a +0.5$.
Considering both possibilities, if the resulting number is odd, it exceeds the maximum number. Therefore, it must be even so we dividing by $2$ getting $1.5p+0.75a+0.25$. However, comparing the inequality $1.5p+0.75a+0.25 < 3p$ to the fact that $a < p$, we find that a point exists between two local minimums, contradicting our assumption of the existence of a cycle other than the $4-2-1$ loop.
It still not proving that there no sequence that goes to infinity though
Please tell me what you think about it.
I don't know if it can help you with your reasoning but I had started building the following graph that represents odd numbers of the sequence with some jumps so that you can have a cyclical representation.
As you can see for each line are represented in the circles the different residue classes of $n \pmod {2^c}$, with increasing exponent $c$.
I have not completed the graph because probably, even if for each branch it is possible to find an increasing pattern, it is impossible to complete it since there will always be a new residue class that does not follow the previous pattern.
However, as you can see, apart from the number $n=1$ followed by $1$ in a cyclical manner, all the other numbers have a smaller number as their next number, therefore the local minimum decreases. In any case, this proves nothing because it is not said that the graph can be completed.