Direct proof of every prime ideal of finite commutative ring with unity is maximal

The usual proof of "every prime ideal of a finite commutative ring with unity is maximal" is pretty straightforward.

$$I$$ is a prime ideal, so $$R/I$$ is a finite integral domain, so $$R/I$$ is a field, so $$I$$ is maximal.

But, how can we prove it without a quotient ring? or how can we prove it directly?

In order to illustrate what I mean by direct proof, let's look at the following example:

We can prove maximal ideals of a commutative ring with unity are prime as follows:

$$I$$ is maximal, so $$R/I$$ a is field, so $$R/I$$ is an integral domain, so $$I$$ is prime.

This approach is not direct, instead, you can check the direct approach here: https://math.stackexchange.com/q/68493

So, I want similar direct proof for my question.

let $$I$$ a prime ideal in a finite commutative ring with unity. We assume that exists an ideal J s.t. $$I\subset J\subset R$$ and let $$a\in J \backslash I$$. because of the limitedness of R there exists $$n,m\in\mathbb N$$ s.t. $$a^n=a^{n+m}\implies a^n(1-a^m)=0\in I.$$ because I is prime we have that $$a^n\in I(\implies a\in I, \text{ contradiction})$$ or $$1-a^m\in I (\implies (a)^m+(1-a^m)=1\in J=R)$$.