That in a ring that in a ring $R$ if the Jacobson radical is maximal ideal, then the ring $R$ is local?
I believe this is true because Jacobson Radical is intersection of maximal left ideals(hence it contains two sided maximal ideals) and hence the Jacobson Radical is contained in every maximal ideal. Because of maximality of Jacobson Radical, we have that any maximal ideal is $J(R)$ and thus $R$ has a unique maximal ideal. Hence $R$ is a local ring.
"The Jacobson radical of $R$ is a maximal ideal iff $R$ is local" is a true statement for commutative rings for the reasons you gave.
In noncommutative rings it is not: take for example $M_2(\mathbb R)$. The Jacobson radical is $\{0\}$, but there are uncountably many maximal left ideals.
Maybe an underlying problem in your question is whether or not you realize the standard notion of "local ring" means "has a unique maximal right ideal" (which is equivalent to the left-handed version.) A ring can have a unique maximal ideal without having a unique maximal left ideal, as the above example shows.
At any rate it is true that if $J(R)$ is a maximal ideal, then it has to be the only maximal ideal. It is contained in (but not always equal to) the intersection of maximal two-sided ideals of $R$, since such ideals are also right primitive ideals.
