Nilpotent or non-Nilpotent Jacobson Radical 
Let $R$ be a ring with identity element such that every ideal of which is idempotent or nilpotent. Is it true that the Jacobson radical $J(R)$ of $R$ is nilpotent?

If $R$ is Noetherian and $J(R)$ is idempotent the Nakayama lemma yields $J(R)=0$. So, for Noetherian rings whose Jacobson radicals are nonzero we have an affirmative answer to the raised question.
Thanks for any help or suggestion!
 A: Let $a$ be an element of $R$. If $\left<a\right>=\left<a\right>^2$, then it is clear that the exists $e^2=e\in \left<a\right>$ such that $\left<a\right>=\left<e\right>$. Now, if $0\not= a\in J (R)$, then $\left<a\right>$ is nilpotent or idempotent. If $\left<a\right>$ is idempotent, by above argument  there exists $e^2=e\in\left<a\right>$ such that $\left<a\right>=\left<e\right>$. Since $e\in J (R)$ , $1-e$ is a unit idempotent and so $1-e=1$. Thus, $\left<a\right>=0$, a contradiction. Therefore, every element of $J (R)$ in nilpotent and so $J (R)$ is nil. If $J (R)$ is finitely generated $J (R)$ is nilpotent.
A: As noted in Rostami's answer, you do get that $J(R)$ is nil, hence is nilpotent if it is finitely generated.  (Here is an alternative proof, just for fun.  Since idempotents are locally 0 or 1, these assumptions imply $J(R) = $nil$(R)$ holds locally, hence globally.)
But here is the main point of my answer:  an example of a quasilocal ring whose maximal ideal is idempotent and all other proper ideals are nilpotent.  So the answer to your question is "no".
(Edited with parenthetical remarks justifying the claims, as requested.)
Let $K$ be a field, $D := K[\{X^s \mid s \in \mathbb{Q}^+\}]$, $M$ be the maximal ideal consisting of elements with zero constant term, and $\overline D_M := D_M/(X)_M$.  Note that the nonzero elements of $D_M$ are each a unit multiple of a power of $X$.  (Given $f \in D$, factor out the biggest power of $X$ that you can to get $f = X^sf_0$, where $f_0 \notin M$.  Since elements of $D$ that are not in $M$ are units in $D_M$, any element of $D_M$ with a numerator of $f$ is a unit multiple of $X^s$.)  Thus $M_M = M_M^2$, hence $\overline M_M = \overline M_M^2$.  If $I_M$ is any other nonzero proper ideal of $D_M$, then there is a positive lower bound on the powers of $X$ it contains, hence $\overline I_M$ is nilpotent.  (If $I_M$ is a proper ideal and there is no positive lower bound on the powers of $X$ that $I_M$ contains, then for each positive rational $s$, there is a $t < s$ with $X^t \in I_M$, hence $X^s = X^{s-t}X^t \in I_M$.  So $I_M$ would contain every power of $X$, hence $I_M = M_M$.)
