Find the rings such that $I^s=\{0\}$ for any maximal ideal $I$ where $s$ is the characteristic. 
Find the rings such that $I^s=\{0\}$ for any maximal ideal  $I$ where
  $s$ is the characteristic.

For any subset $A$ of a ring and any positive integer $s$, deinfe $A^s=\{a^s \mid a\in A\}$, where $a^s=\underbrace{a\cdot a \cdot \cdots \cdot a}_{\text{$s$ times}}$.
I have investigate the following examples.


*

*For finite field $\mathbb{F}_q$, the only maximal ideal is $\{0\}$. It holds.

*For residual class ring $\mathbb{Z}_n$ with $n=\prod_{i=1}^{k}p_i^{e_i}$ where $p_i$ are primes and $e_i$ are positive integers, the maximal ideals $I_i$ are principle and generated by $p_i$, namely $I_i=(p_i)$ for $i=1,2,\cdots,k$. If $k=1$, then it holds since $n\mid p_1^n$. If $k>1$, then it does not hold since $p_j\nmid p_i^n$ for $1\le i\ne j\le k$.


More rings can be discussed, but I want the condition applicable to most rings. Does it require the ring has only one maximal ideal ?
 A: Since the Jacobson radical contains nilpotent ideals, and a maximal ideal is nilpotent, it follows that the Jacobson radical is maximal and therefore the ring is local. (Including this for completeness, although it has already been pointed out by Andrew and a commenter.)
If, furthermore, we require $J(R)^s=\{0\}$ where $s>0$ is the characteristic, then there is an additional point. A local ring of positive characteristic $n$ necessarily has $n=p^k$ for some prime $p$, since a local ring only has trivial idempotents.
So we can say that a ring $R$ having your properties is characterized by the three things:


*

*it has positive characteristic

*it is local with a nilpotent Jacobson radical (a special sort of semiprimary ring) 

*The index of nilpotency of the radical is limited ($\leq \mathrm{char}(R)$.


I can't think of any other characterization that reduces the nilpotency of the Jacobson radical to something simpler. Usually this is taken as a condition to be studied in itself, and this would not be done if there were simpler characterizations.
You can make such a ring out of any local ring of positive characteristic simply by modding out by $M^n$ where $M$ is the unique maximal ideal and $1\leq n\leq \mathrm{char}(R)$.

An ideal $$ is nilpotent if there is $$ such that $i=0$ for all $∈$. How to say $=$ ? – zongxiang yi

That is not the definition of a nilpotent ideal, it is the definition of a nil ideal. A nilpotent ideal satisfies the stronger requirement that each product $i_1\cdots i_n=0$ for any collection of elements $i_j\in I$.  It will usually not be the case that all elements have the same degree of nilpotency.
A: The Jacobson radical of a (not necessarily commutative) ring is the intersection of all maximal left (or all maximal right) ideals, and contains every nilpotent ideal.
So, in your situation where every maximal left ideal is nilpotent, it follows that there is a unique maximal left ideal. Thus every such ring is local with nilpotent radical (so local and semiprimary).
