We suppose $0<m<n$. It is known, from Landau's paper quoted above that
$
n\to n^{1/3}J_n(x_n)
$
increases (to some universal constant $b$). Thus
$$
J_n(x_n)>\left(\frac{m}{n}\right)^{1/3}J_m(x_m)>\left(\frac{m}{x_n}\right)^{1/3}J_m(x_m).
$$
It is also known that for $x>n$,
$
x\to \sqrt{x^2-n^2}((J_n(x))^2+(Y_n(x))^2)
$
increases to $2/\pi$. Therefore, as $x_n>n>m$
$$
|J_m(x_n)|^2\leq \frac{2}{\pi}\frac{1}{\sqrt{x_n^2 -m^2}}.
$$
Therefore, the inequality holds at least for all $n,m$ such that
$$
\frac{2}{\pi m^{2/3}(J_m(x_m))^2}<\sqrt{x_n^{2} -m^{2}} \frac{1}{x_n^{2/3}},
$$
and the left hand-side is between $1.39$ and $1.89$.
If we write $$m=\frac{x_n}{1+\lambda x_n^{-2/3}}$$ we find that this inequality holds for all $x_n\geq40$, and $\lambda>11/10$ for example. The asymptotic of $x_n$ is quite well known, and it is precisely of that sort of magnitude, $x_n=n+\beta n^{1/3}+O(1)$. So, refining the argument and using $n$ instead of $x_n$, you would be very close. For $m=n-1$, you get
$$
\frac{2}{\pi m^{2/3}(J_m(x_m))^2}<\sqrt{\beta^2 n^{2/3} +2\beta n^{4/3}+2n-1} \frac{1}{n^{2/3}},
$$
Asymptotically, this is not enough, because $\beta\approx 0.8$ and $\sqrt{1.6}=1.26<1.39$, but that does limit the range of $m$ to investigate, to $m\geq n- n^{1/3}/5$, so you are only have to consider the case when $J_m(x_n)>0$, $J_m^\prime(x_n)<0$, as the first root is further away ($m+ 1.8m^{1/3}$).