Someone asked a question on Math exchange yesterday, and I showed in my reply that the smallest element (a_0 ) of a cycle has the property that a_0<(\frac{3}{2})^k (translated to your notation of k and n which are inverted here ,see https://math.stackexchange.com/questions/5085433/explicit-baker-constants-for-collatz-cycle-constraints/5085630#5085630)
This would place the largest element at a_{max}<(\frac{9}{4})^k. If we start using tighter bounds (like for the 2.56^k bound) in the logic, we could even have something like a_0<(\frac{3}{2.5})^k or a_{max}<(\frac{9}{5})^k which would place all elements of the cycle a_i<2^k<3^k<2^n
Still not elementary but might be useful