Collatz Conjecture: Understanding why for a given $K, L$, there are only a finite number of solutions

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I am reading through this paper by
by Simon & Weger regarding the Collatz Conjecture.

I am stuck on the reasoning at the end of 2.2

Let:

  • $n$ be a natural number.

  • $T(n) = \begin{cases} \frac{1}{2}(3n + 1), && \text{if }n\text{ is odd}\\ \frac{1}{2}n, && \text{if }n\text{ is even}\\ \end{cases}$

  • sequence be an increasing subsequence of odd integers followed by a decreasing subsequence of even integers

  • a cycle be an $m$-cycle if it consists of $m$ sequences with a total of $K$ odd numbers and a total of $L$ even numbers.

  • a sequence is periodic if there exists an integer $p \ge 1$ in the sequence $\{ n, T(n), T^2(n), \dots, T^{p}(n) \}$ where:

    • $T^0(n) = n$
    • $T^{i+1}(n) = T(T^i(n))$
    • $T^p(n) = n$
  • $t_0, t_1, \dots, t_{m-1}$ be the indices of the $m$ local minima in an $m$-cycle such that:

    • $t_0 = 0$
    • $t_0 < t_1 < t_2 < \dots < t_{m-1} < p$
  • $s_0, s_1, \dots, s_{m-1}$ be the indices of the $m$ local maxima in an $m$-cycle such that:

    • $t_0 < s_0 < t_1 < s_1 < \dots < t_{m-1} < s_{m-1} \le p-1$
  • $x_i, y_i$ be the values of the local minima and maxima so that:

    • $x_i = T^{t_i}(n)$
    • $y_i = T^{s_i}(n)$
  • $k_i, l_i$ be defined so that:

    • $k_i = s_i - t_i$ for $i = 0, \dots, m-1$
    • $l_i = t_{i+1} - s_i$ for $i = 0, \dots, m-2$ and $l_{m-1} = p + t_0 - s_{m-1}$
    • $K = \sum\limits_{i=0}^{m-1}k_i$
    • $L = \sum\limits_{i=0}^{m-1}l_i$
  • $M = \begin{pmatrix} -3^{k_0} & 2^{k_1 + l_0} & 0 & \cdots & 0\\ 0 & -3^{k_1} & 2^{k_2 + l_1} & \ddots & \vdots\\ \vdots & \ddots & \ddots & \ddots & 0\\ 0 & 0 & \ddots & -3^{k_{m-2}} & 2^{k_{m-1} + l_{m-2}}\\ 2^{k_0 + l_{m-1}} & 0 & \cdots & 0 & -3^{k_{m-1}} \end{pmatrix}$

  • $\Delta = 2^{K+L} - 3^K$

  • $j* = \begin{cases} j, && \text{if }i \le j\\ j+m, && \text{if }i > j\\ \end{cases}$

  • $\Delta M^{-1} = (m_{i,j})_{i,j}=0,1,\dots,m-1$

  • $\alpha_{i,j} = \sum\limits_{h=i+1}^{j*}k_h$

  • $\beta_{i,j} = \sum\limits_{h=i+1}^{j*}l_{h-1}$

It follows that:

  • $M\begin{pmatrix} a_0\\ a_1\\ \vdots\\ a_{m-1} \end{pmatrix} = \begin{pmatrix} 2^{l_0}-1\\ 2^{l_1}-1\\ \vdots\\ 2^{l_{m-1}}-1 \end{pmatrix}$

  • $\det(M) = (-1)^{m-1}\Delta$

  • $m_{i,j} = 2^{\alpha_{i,j}+\beta_{i,j}}3^{K-k_i-\alpha_{i,j}}$

  • $m_{i,j} > 0$

  • $\Delta > 0$

  • $m$-cycles are in one-to-one correspondence with solutions $k_i, l_i, a_i$ of the matrix equation with $k_i, l_i, a_i$ are all positive integers.

Here is the part that I am not clear on:

  • It follows that when we know upper bounds for $K$ and $L$, there are only finitely many rational solutions.

I am trying hard to understand why the above establishes that there are only finitely many solutions given $K$ and $L$.