Is there a completely elementary way to prove that $Y^2=X^3-32X$ has rank 1?

I’m working on a paper in which I end up considering the biquadratic rational curve $$u^2v^2 - u^2 - v^2 - 6uv + 8 = 0. \tag{$1$}$$ To complete the remainder of my proof/method, I need to prove that it has rank 1. I believe it can be transformed to the Weierstrass form $$Y^2=X^3-32X,$$ and then Sage gives

Weierstrass Form: Elliptic Curve defined by y^2 = x^3 - 32*x over Rational Field
Generators: [(-4 : 8 : 1)]
Rank: 1
Torsion Points: [(0 : 1 : 0), (0 : 0 : 1)]

Since my paper is completely elementary other than this single step, I’m hoping there’s a way to prove that (1) has rank 1 in an elementary way. At the very least, I’m hoping there’s a relatively short and accessible way to avoid the “black box” of Sage computation, and possibly write out the calculation “by hand” in the paper.

Any suggestions/advice would be appreciated!

Note: I asked an LLM, and it said “look for rational numbers $d$ (modulo squares) that make the following equation solvable in integers $N^2=dM^4-\frac{2}{d}e^4$ where $d \mid -2$”, then showed that those four equations were solvable so the “image size” is 4, then did the same for the “partner curve” $\hat{E}: y^2=x^3+8x$ which apparently had image size $2$, and then used $2^r = (4x2)/4 = 2$ to deduce $r=1$.

I have no idea if that is both correct and “elementary”, but at least I understand it, and that — or something like it (but correct!) — would certainly be “elementary” enough for me to happily include in my paper.