Let $R=\left( \begin{array}{ccc} \mathbb{Z}&0\\ \mathbb{Q}&\mathbb{Q} \end{array}\right)$.
It is known to be left hereditary but not right hereditary. But I don't know how to prove it. Please give me some hints or answers.
Small’s example of left hereditary ring but not right hereditary
abstract-algebraring-theory
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Yes, your reasoning is valid that if $M_n(I)\subsetneq M_n(K) \subseteq M_n(R)$, then $I\subsetneq K\subseteq R$, and so if $I$ is maximal, $M_n(I)$ is maximal.
As for the title question, a maximal ideal in a noncommutative ring is always prime, unless you mean to apply the commutative definition of prime ideals to noncommutative rings. The general definition of "prime ideal" is "$AB\subseteq P\implies A\subseteq P \text{ or } B\subseteq P$."
In noncommutative algebra, an ideal that satisfies $ab\in I\implies a\in I \text{ or } b\in I$ is called a completely prime ideal, and the zero ideal of a matrix ring over a field is an example of a maximal ideal that isn't completely prime.
The example you gave is a maximal and prime ideal of $M_n(\Bbb Z)$ which is not a completely prime ideal.
Yes, I think you translated the condition correctly. The easiest way for me is to remember this ring is anti-isomorphic (via transposition) to the one in the link you gave.
If $I_1=\{0\}$, then $I_2$ can be any submodule of $K\oplus K$.
If $I_1=K$, then $I_2$ has to contain $\{0\}\oplus K$, so it has to be of the form $K\oplus K$ or $\{0\}\oplus K$.
These last two correspond to
$$\begin{pmatrix} K & 0\\ K & 0 \end{pmatrix}$$
and
$$\begin{pmatrix} K & 0\\ K & K \end{pmatrix}$$
and the ones of the first form look like
$$L_W=\left\{\begin{pmatrix} 0 & 0\\ b & c \end{pmatrix}\middle|\, (b,c)\in W\right\}$$ for a given subspace $W< V$.
Note that if you pick any $1$-dimensional subspace $W$, that is automatically going to make a minimal left ideal. The first two in the list are obviously not minimal because they contain the left ideal of strictly lower triangular matrices.
We can show all minimal left ideals are isomorphic to $L_{K\oplus 0}$. Suppose $(a,b)$ is a nonzero element of $L_W$ where $W$ is one dimensional. If $b=0$ then obviously there is nothing to do. If $b\neq 0$, then
$\begin{pmatrix}0&0 \\ b^{-1}&0\end{pmatrix}$ defines a left $R$ module transformation from $L_W\to L_{K\oplus 0}$ by right multiplication, which is an isomorphism because they're both one dimensional.
So look at what you have:
$$\begin{pmatrix} K & 0\\ K & K \end{pmatrix}\cong \begin{pmatrix} K & 0\\ K & 0 \end{pmatrix}\oplus \begin{pmatrix} 0 & 0\\ 0 & K \end{pmatrix}$$, so both of these pieces are projective, being summands of a free left module.
Then, all of the $L_W$ with one-dimensional $W$ are isomorphic to the second factor, so they are projective as well. If $W$ is two-dimensional, then $L_W=\begin{pmatrix} 0 & 0\\ K & K \end{pmatrix}$ which is obviously a direct sum of two minimal left ideals, and thus projective (since they are.)
So, all left ideals are projective.
Best Answer
A correction: this ring is left hereditary and not right hereditary.
It has been covered here how to show that the strictly triangular subring is a one-sided ideal that isn't projective, which would prove that it is not right hereditary (after you adapt it for the side change.)
A proof appears on page 46 of Lectures on rings and modules by T.Y. Lam. Using the opposite ring instead of your version. The strategy used is to use the classification of right ideals (left ideals in your case) to show that they are all projective.
It turns out that there are three types of left ideals:
$A_n=\begin{bmatrix}n\mathbb Z&0 \\ \mathbb Q&\mathbb Q\end{bmatrix}$
$B_n=\begin{bmatrix}n\mathbb Z&0 \\ \mathbb Q&0\end{bmatrix}$
$C_V=\left\{\begin{bmatrix}0&0 \\ p&q\end{bmatrix}\middle |(p,q)\in V, V \text{ a subspace of $\mathbb Q^2$}\right\}$
The types $A_n$, $n\neq 0$ are all free, and because they split into $A_n\cong B_n\oplus\begin{bmatrix}0&0\\ 0& \mathbb Q\end{bmatrix}$, the $B_n$ are also projective.
The trickiest part is to show the ones of the form $C_V$ are projective, and I don't have enough time at the moment to give a complete argument, but I will when I can. Until then I hope the reference and the clue will get you on your way.