First off, I would not like to use any facts about quotient rings. I've got a proof but I need further explanations for some steps.
Theorem Let $R$ be a commutative ring with unity. Then any maximal ideal of $R$ is prime.
Proof: Let $M$ be maximal ideal of $R$. For the sake of contradiction, suppose $M$ is not prime. That is there are some $a,b\in R$ s.t. $ab\in M$ and $a,b\notin M$. Consider the ideal $(a)+M$. We have that $(a)+M$ is strictly larger than $M$ thus $R=(a)+M$ by maximality of $M$. Since $R$ is unitary, we have $1\in R=(a)+M$ so $$1=ra+m$$ for $r\in R,m\in M$ similarly, we have $R=(b)+M$ and $$1=sb+n$$ for some $s\in R, n\in M$ From these equalities, we obtain \begin{align*} 1&=1\cdot1\\ &=(ra+m)(sb+n)\\ &=rsab+ran+msb+mn \end{align*} Since $ab,m,n$ are all in $M$ the last expression is in $M$. This yields that $1\in M$ so $M=R$. Since maximal ideal is a proper ideal by definition, this is a contradiction. Thus, R must be prime.
Now, I am quite unsure about some steps, what exactly means the ideal $(a)+M$? Is it a ideal of the form $\{ra+m,r\in R,m\in M\}$? That's what im getting, right? Then, why should the expression $rsab+ran+msb+mn$ be in $M$? EDIT: Also, why should $1\in M$ imply that $M=R$?
$ab\in M$ implies that $rs(ab)\in M$ since $M$ is an ideal, $n,m\in M$ implies that $(ra)n, (sb)m, mn\in M$ since $M$ is an ideal, we deduce that $rsab+ran+msb+mn=rs(ab)+ran+sbm+nm$ is a sum of elements of $M$.