Collatz conjecture but with $n^2-1$ instead of $3n+1.$ Does the sequence starting with $13$ go to infinity? Let's consider the following variant of Collatz $(3n+1) : $
If $n$ is odd then $n \to n^2-1.$
$1\to 0.$
$3\to 8\to 1\to 0.$
$5\to 24\to 3\to 0.$
$7\to 48\to 3\to 0.$
$9\to 80\to 5\to 0.$
$11\to 120\to 15\to 224\to 7\to 0.$
$\color{red}{13\to 168\to 21\to 440\to 55\to 3024\to 189\to 35720\to 4465\to 19936224\to 623007\to\ldots\ ?}\ $

Does the sequence starting with $13$ go off to infinity? If yes, what
is a proof? If no, is there a  starting number whose sequence does go off to infinity, and
how do we prove either that such a number must exist, or even better that a specific starting number goes off to infinity?

Here is some Python code I ran, which suggests that the numbers in the sequence starting at $13$ quickly become large:
n=13
num_loops=0
print(n)
while n!=0:
    if n%2==0:
        n//=2
    else: n=n**2-1
    print('\n', n)
    num_loops+=1
    if num_loops==70:
        print("too many loops")
        break

 A: About Eric Syder's comment:  suppose $   x^2 - 1 = 2^k y$  with $x,y$ odd.  We wish to investigate what happens when $y \leq x \; . \; \;$  Note that $\gcd(x+1, x-1)  = 2$  because $x \equiv 1,3 \pmod 4.$  One of the $x \pm  1 $ is $\equiv 2 \pmod 4. $
Let us make the name $  \delta = \pm 1.$    Then we may demand
$$ x + \delta \equiv 2 \pmod 4$$
This tells us that the integer $ \frac{x + \delta}{2} $  is odd. We also have
$$      \frac{x + \delta}{2}   \frac{x - \delta}{2} = 2^{k-2}y$$
By repeated division by $2$   it follows that
$$ \frac{x - \delta}{2^{k-1}} = w $$   is an odd positive integer,  with
$$\frac{x + \delta}{2} \; \frac{x - \delta}{2^{k-1}} = y $$
IF  WE ASSUME $$ w = \frac{x - \delta}{2^{k-1}}  \geq 3,  $$
we find
$$   y = w \frac{x + \delta}{2}  \geq 3 \frac{x + \delta}{2} \geq \frac{3x - 3}{2}$$
The assumption $ w \neq 1$  has led us to $y \geq \frac{3x - 3}{2}$  The hypothesis that $x \geq y$  now says  $x \geq \frac{3x - 3}{2},$   or
$ 2x \geq 3x - 3,$ or $3 \geq x$
If $ x \geq y$   and $  x \geq 5$  in $   x^2 - 1 = 2^k y,$   we find that $w=1$  in $ \frac{x - \delta}{2^{k-1}} = w .$   So that $ x - \delta = 2^{k-1} .$  or
$$   x = \pm 1 + 2^{k=1}$$
