Right. And the irrationality measure of your $x$ is $3$ (and not $\infty$ as I had previously commented in a hurry).

For the record (and for those interested), the result mentioned by Noam in the comments above (as for the irrationality measure of $e^{2/k}$ when $k$ is a non-zero integer) is due to C. S. Davis, see Rational approximations to $e$, J. Austral. Math. Soc. Ser. A 25 (1978), 497-502.

Yes, that's it. But now I wonder if I should have raised Q2 in a separate thread. In any case, thank you much for the fruitful exchange.

If $XY \ne YX$, then $X^mY=YX^m$ is impossible for $m \ge 1$. For suppose wlog $XY \prec YX$. Then, $X^m Y \prec X^{m-1} Y X \prec \cdots \prec XYX^{m-1} \prec YX^m$, which is the same argument used in the OP to prove that ${\rm BS}(n,n)$ is not bi-orderable for $n \ge 2$ (it works as well for $|n| \ge 2$).

As for your 2nd comment: $[X^m, Y] = 1$ iff $X^m Y = Y X^m$, and this is impossible if $m \ge 2$ for the same reason expressed in the OP with reference to the orderability of ${\rm BS}(n,n)$ when $n \ge 2$.

Many thanks for the interest, Yves. Just a minor detail as for your 1st comment: I think it should be $Y = y^q$, so $X^s = Y^s = 1$ isn't true. Nevertheless, $X^s=Y^s$ is enough to conclude: Since $X\ne Y$ (here is where we use that $s =\gcd(m,n) \ge 2$), then either $X\prec Y$, and then $X^s\prec Y^s$, or $Y \prec X$, and then $Y^s\prec X^s$.

Right, I was too hasty (and optimistic). But then, what if $z = 2x$? Actually, this is the case I am interested in (I edited the OP according to your comment).