An ideal $P$ of a ring $R$ is called prime (In ideal sense) if for ideals $I,J $ of $R$, $P$ contains the product ideal $IJ$ implies $P$ contains $I$ or $P$ contains $J$.
At this moment we call prime ( in ideal sense) as (ideally) prime.
I can't found a ring and one maximal ideal which is not (ideally) prime but also cannot disprove the exsitance of such an animal.
What I've got is If such an animal exists, it must lies in a non commutative ring, or a non unital ring (actually stronger, $R^2$ is strictly contained in $R$)
Thank you.
By the way, I have an example that an ideal is maximal but not prime in the usual sense ( for any two elements $a,b $ in $R$, $ab$ in $P$ implies $ a$ in $P $ or $b $ in $P$ ), unfortunately that non prime ideal is Ideally prime. It is the zero ideal in the matrix ring with entry in an division ring.
When $R$ is unital, the answer is no.
Let $P$ be a maximal ideal, and choose any two ideals $I$ and $J$ such that $I\not\subseteq P$ and $J\not\subseteq P$. Then $P+I$ and $P+J$ are ideals of $R$ that contain $P$ and some stuff not in $P$ - and so they're bigger than $P$, so (as $P$ was maximal) they must be equal to $R$.
In particular, we can write $1\in R = P+I$ as $1 = p+i$ for some $p\in P$ and $i\in I$, and $1\in R = P+J$ as $1 = p'+j$ for some $p'\in P$ and $j\in J$. Multiplying these two expressions together, we get $$1 = (p+i)(p'+j) = pp' + ip' + pj + ij.$$ But it's obvious that $pp'$, $ip'$ and $pj$ are in $P$. Hence $ij\in P$ if and only if $1\in P$. But we know already that $1\not\in P$, because $P$ is a maximal ideal (and hence by definition $P\neq R$). So $ij\not\in P$, and hence $IJ\not\subseteq P$.
When $R$ is non-unital, the answer is yes. For example, the non-unital ring $R = 2\mathbb{Z}$ contains the maximal ideal $P = 4\mathbb{Z}$, and the ideals $I = J = 2\mathbb{Z}$. Then $IJ\subseteq P$, but $I, J\not\subseteq P$.