Diophantine equation $2(x - 1/x) = y - 1/y$ Does the Diophantine equation $2(x - \frac{1}{x}) = y - \frac{1}{y}$ have only trivial rational solutions, i.e, $x=\pm1, y = \pm1$?

This question is actually the first example, in a different formulation, of the following problem:
Find two integer right-angled triangles with a common base and altitudes in the integer ratio $N:1$, 
which was considered by the late John Leech (in the 1980s I think).
We have to find integer solutions to
$B^2+A^2=C^2 \hspace{2cm} B^2+(NA)^2=D^2$
We can express $B=\alpha  2mn , A = \alpha (m^2-n^2)$ and $B=\beta 2pq , NA=\beta (p^2-q^2)$ giving
$\frac{NA}{B}=\frac{p^2-q^2}{2pq}=N\frac{A}{B}=N\frac{m^2-n^2}{2mn}$
and defining $X=m/n$ and $Y=p/q$ gives
$N(X-1/X )=(Y-1/Y)$
The original form of the problem is also just that of Euler's Concordant Numbers which is related to the
elliptic curve
$V^2=U(U+1)(U+N^2)$
which has $3$ torsion points of order $2$ when $U=0,-1,-N^2$ and $4$ torsion points of order $4$ when $U=\pm N$,
all of which give the trivial answer or lead to undefined quantities.
The first curve of rank greater than $0$ occurs for $N=7$ leading to $X=3/2, Y=6$ or $B=12, A=5$.
