There are no infinite descending chains, so the powers of J stabilize at some ideal H. Suppose H is nonzero. Consider the family of left ideals U where HU is nonzero. Here HU is the product of the two left ideals. When U = H the result is H, so our family is nonempty.
Start with any such U and find a smaller ideal in the family, then a smaller one, until you reach the end of the finite chain. In other words, we may assume U is minimal with respect to HU ≠ 0.
Choose an element c in U with Hc nonzero. Then multiply Hc by H on the left. Since H is two sided, we can invoke associativity, hence H(Hc) = (HH)c = Hc. Thus Hc is a left ideal inside U with HHc nonzero. Since U is minimal, U = Hc for some c.
What does the product Hc look like? Since c is a single element of R, Hc is merely the elements of H scaled by c on the right. This is already a left ideal. We don't have to take finite sums, or anything inconvenient like that.
To recap, U = HU = HHc = Hc.
Since c is in U, there is some y in H satisfying c = yc. Thus (1-y)c = 0. However, y is in J, so 1-y is a unit, and c = 0. This is a contradiction, so H = 0, and J is nilpotent. The jacobson radical of a left artinian ring is nilpotent.
Remember that J contains all nil and nilpotent ideals. Thus J is the largest nilpotent ideal, and all nil ideals are nilpotent, with exponent bounded by the exponent of J.
In contrast, let R be noetherian, a dvr for example. Let x generate an ideal, a power of the maximal ideal M. Since powers of this ideal are never 0, x is not nilpotent. There are no nilpotent elements, and no nil ideals.