= Solution
The descending powers $J\supseteq J^2\supseteq\cdots$ stabilize because they are <right ideals>. Choose $m\ge1$ with $J^m=J^{m+1}$ and put $I=J^m$. Then $I=J^{2m}=I^2$.
Suppose $I\ne0$. Among the <right ideals> $K$ such that $KI\ne0$, choose a minimal one; the set is nonempty since $RI=I\ne0$. The <right ideal> $KI$ lies in $K$, and $(KI)I=KI^2=KI\ne0$, so minimality forces $KI=K$. Choose $k\in K$ with $kI\ne0$. Then $(kR)I=kI\ne0$, and $kR\subseteq K$, so the same minimality gives $K=kR$.
Since $K=KI$ and $I\subseteq J$, its generator satisfies $k=\sum_\nu kr_\nu i_\nu=kj$ for an element $j=\sum_\nu r_\nu i_\nu\in J$. Thus $k(1-j)=0$. The <unit criterion for the Jacobson radical> makes $1-j$ invertible, implying $k=0$, contrary to $kI\ne0$. Therefore $I=0$, and
$$
\boxed{J(R)^m=0.}
$$
The <minimal-right-ideal proof of radical nilpotence> is important here: one cannot apply the finite-generator form of <Nakayama lemma> directly to $J^m$ before proving it finitely generated. We made a suitable auxiliary <right ideal> cyclic before using the unit argument.
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