Assuming 2^n > 3^k (i.e. non-negative cycles)
We know that the maximal element of a circuit with n elements and k odds is 2^{n-k}\frac{3^k-2^k}{2^n-3^k}.
We know that it is also bigger than any element of any non-negative Collatz cycle with n elements and k odds (@mathkook could you confirm this is true and that the proof is simple? Ideally do you have the proof and/or a ref?)
In the thread Elementary proof of 2^n > 2^(n-k) (3^k - 2^k) / (2^n - 3^k) provided 2^n > 3^k - #4 by cosmo I give a draft, elementary, proof that:
2^{n-k}\frac{3^k-2^k}{2^n-3^k} < 2^n
EDIT: the proof was incorrect 
In turns, this gives an elementary proof that any elemnt of any non-negative Collatz with n elements is less than 2^n.