I'm not a mathematician but this problem interested me so I thought about it for a bit. My current understanding is that it needs proof that there is NOT a number that fails to pass the conjecture, but what if we simply prove that $3x+1$ DOES cycle all factors?
$4,2,1$ in factors is just just ($2 \cdot 2$), ($2$), ($1$). Since $1$ is an odd number, apply $3x+1$ which means $3(1)+1=4$, or ($2\cdot 2$), and it loops. For any odd number, $x$, the expression $3x+1$ always results in creating $2$ as a factor, which we then remove by $n/2$. The end result is the original number has had its factors shuffled.
For example, if $x=9$, then
$$3(9)+1 = 3(3 \cdot 3)+1 = 28 = (7 \cdot 2 \cdot 2)$$
So we then eliminate the $2$s and result in a transformation of the factors from ($3\cdot 3$) to just ($7$).
I would hypothesize that shuffling like this eventually will always result in a permutation that contains only $2$ to some power as factors. If in theory we can prove that $3x+1$ cycles through all prime numbers, doesn't that prove that eventually you will invariably end up with a number that contains only factors of $2$? Or is that proof not good enough? Also I don't know how to prove that, obviously. I just had this idea, tried looking it up and found nothing so I wasn't sure if people hadn't thought about it or if there is a good reason why this isn't considered.
Thank you for your time.