In the "syracuse"-formulation of the collatz-transformation $T:= a_{n+1}={3a_n+1 \over 2^{A_n}}$ on odd a where $A_n$ is the greatest value that keeps $a_{n+1}$ integer we can eleganly write a formula to approach the cycle-problem.
Let's write in the following example $a,b,c,...$ for $a_1,a_2,a_3,...$ and $A,B,C,...$ for $A_1,A_2,A_3,...$ then we have for the example-assumtion of a 3-step-cycle
$$ b= {3a+1\over2^A} \qquad c= {3b+1\over2^B} \qquad a= {3c+1\over2^C} \qquad $$ and the equation must hold:
$$ bca = \left({3a+1\over2^A}\right)\left({3b+1\over2^B}\right)\left({3c+1\over2^C}\right)$$
Reordering gives the "critical equation" for the general cycle problem (it is obvious how it can be extended to as many assumed steps as wished?) with $S=A+B+C+...$ and $N=(number-of-steps)$
$$ 2^S = \left(3+{1\over a}\right)\left(3+{1\over b}\right)\left(3+{1\over c}\right) \\ =3^N \left(1+{1\over 3a}\right)\left(1+{1\over 3b}\right)\left(1+{1\over 3c}\right) $$
One can immediately see, that the S must be so that a perfect power of 2 is slightly greater then the N 'th perfect power of 3. But we find empirically, that perfect powers of 2 and 3 are not very near - so the parentheses must be big and thus the elements $a_k$ of the cycle must be small ! You might try to find yourself possible $a,b,c$ for an assumed 3-step cycle which gives the critical equation
$$ 32 = 27 \left(1+{1\over 3a}\right)\left(1+{1\over 3b}\right)\left(1+{1\over 3c}\right) $$
and only assuming the most general facts, that $a,b,c$ are all odd, distinct and not divisible by 3. Can you disprove the 3-step-cycle?
Well, since we know, that all $a_k \lt 2^{50} $ are not member of a cycle, the fractions in the parentheses are very small and we need a lot of parentheses and thus a high N, to increase $3^N$ to the next perfect power $2^S$ - by this formula we can derive a general lower bound for the number of steps required given a lower bound for the elements of the cycle $a_k$ .
This "critical equation" does not suffice for the disproof of cycles alone, because for any lower bound of a we can find a lower bound of N which lets the rhs reach and "overtake" the next perfect power of 2. The next step of possible research seems then to re-introduce the remaining modular requirements on subsequent $a_k$ . There has been real progress on the general "critical equation" assuming certain structures in the $a_k$, namely that of an "1-cycle" and "m-cycle" structure, specific simple and restricted forms of assumed cycles, which could then be disproved by the known distance-characteristic of powers of 2 to powers of 3, using the continued fraction of $\log(3)/ \log(2)$ and theory of rational approximation of linear forms in logarithms.
(See the wikipedia article where I also linked to online available articles of John Simons and Benne de Weger. The basic article of Ray Steiner is unfortunately not freely available online - this is really a grief!)
The answer to your question is yes; one can work the Collatz relation backwards to build numbers that last arbitrarily long (for a simple example, just take powers of 2). However the purpose of the wikipedia records is to see if we can find SMALL numbers that last a long time.
Best Answer
If you proved that Collatz was not disprovable in (say) PA, this would mean that there were no finite cycles not hitting $1$. However, it would leave open the possibility that there was some number $n$ which never hit $1$, and never entered a finite cycle (just "shot off to infinity"); so this wouldn't actually prove the Collatz conjecture.
I believe there have been no serious attempts at proving Collatz to be unprovable in PA (or related theories). The problem is that we only know a small handful of techniques for showing PA-unprovability, and none of them seem to apply to Collatz. Proving unprovability is extremely hard.
Incidentally, the statements for which PA-unprovability implies truth are those which are equivalent in PA to a $\Pi^0_1$ sentence; it is not known whether the Collatz conjecture is equivalent over PA to a $\Pi^0_1$ sentence, so currently no way is known for turning a proof of the PA-undisprovability of Collatz into a proof of the Collatz conjecture.