Find a left ideal of $\mathbb{H}[X]$ that is a maximal, but not a ideal. 
Find a left ideal of $\mathbb{H}[X]$ that is a maximal, but not a ideal.

$\mathbb{H}[X]$ is just the polynomial extension.
$$\mathbb{H}=\{a+bI+cJ+dK \mid a,b,c,d \in \mathbb{R} \} .$$
$$\mathbb{H}[X]=\{\alpha_0+\alpha_1X+\cdots+\alpha_nX^n: \alpha_0,\alpha_1,\dots,\alpha_n\in \mathbb{H}\}$$
First thought is something like $\mathbb{H} \times (XI)$. However, sort of stuck. I know you have to use the non-commutativity of quaternions somewhere.
Second thought is that you have to use something about inverses. Well, it can't have $1$ in it. Since you would get itself. So, ... I'm pretty sure $\mathbb{H} \times (XI)$ is the left ideal. However, can't see how you would get not a right ideal.
 A: New answer
Let $\mathbb H$ be the $\mathbb R$-algebra generated by the symbols $i_0$ and $j_0$ subject to the relations 
$$
i_0^2=-1=j_0^2,\quad i_0\ j_0=-j_0\ i_0.
$$
Let $X$ be an indeterminate and define the $\mathbb R[X]$-algebra $\mathbb H[X]$ by 
$$
\mathbb H[X]:=\mathbb R[X]\underset{\mathbb R}{\otimes}\mathbb H.
$$ 
We must find a maximal left ideal $\mathfrak a$ of $\mathbb H[X]$ which is not an ideal. 
Put 
$$
\mathbb C:=\frac{\mathbb R[X]}{(X^2+1)}
$$
and denote by $x\in\mathbb C$ the canonical image of $X$. 
Note that the left ideal $\mathfrak b\subset\mathbb H[X]$ generated by $X^2+1$ is an ideal. In particular
$$
A:=\frac{\mathbb H[X]}{\mathfrak b}
$$ 
is a $\mathbb C$-algebra. 
If $\pi:\mathbb H[X]\to A$ is the canonical projection, and if $\mathfrak c$ is a maximal left ideal of $\mathbb H[X]$ which is not an ideal, then $\mathfrak a:=\pi^{-1}(\mathfrak c)$ fits the bill. 
Let $M$ be the $\mathbb C$-algebra of two by two matrices with entries in $\mathbb C$, and let $i,j\in A$ be the canonical images of $1\otimes i_0,1\otimes j_0\in\mathbb H[X]$. 
One easily checks that there is a unique $\mathbb C$-algebra morphism $\phi:A\to M$ satisfying 
$$
\phi(i)=\begin{pmatrix}x&0\\0&-x\end{pmatrix},\quad 
\phi(j)=\begin{pmatrix}0&-1\\1&0\end{pmatrix},
$$
and that $\phi$ is bijective. 
It is also easy the verify that the left ideal of $M$ generated by 
$$
e:=\begin{pmatrix}1&0\\0&0\end{pmatrix}=\phi\left(\frac{1-ix}{2}\right)
$$
is maximal, and that it is not an ideal. 
This implies that we can put 
$$
\mathfrak c:=A\ \frac{1-ix}{2}\quad,
$$
or, in other words, that 
$$
\pi^{-1}\left(A\ \frac{1-ix}{2}\right)
$$
is a maximal left ideal of $\mathbb H[X]$ which is not an ideal. 
Old answer
Let $\mathfrak a$ be the ideal of $\mathbb H[X]$ generated by $X^2+1$, let $A$ be the quotient $\mathbb H[X]/\mathfrak a$. 
It suffices to find a maximal left ideal of $A$ which is not an ideal. 
Write $x\in A$ for the canonical image of $X$. Let us identify $\mathbb R[x]$ to $\mathbb C$. Then $A$ is isomorphic, as a $\mathbb C$-algebra, to $M_2(\mathbb C)$. 
(See the very beginning of the text of Gaëtan Chenevier named Lecture 6 on this html page. Here is a direct link to the pdf file.)
As a result, $A$ contains an idempotent $e$ with $1\neq e\neq0$, and $Ae$ is the sought-for maximal left ideal.
EDIT. One can put 
$$
e:=\frac{1+ix}{2}\quad.
$$
(I use the traditional notation $i,j,k$ instead of the $I,J,K$ in the question.)
A: I am assuming that in the polynomial ring $R=\mathbb{H}[X]$ the indeterminate $X$ commutes with all the constants, for otherwise what is $Xq$ for some $q\in\mathbb{H}$?
Unless I am very wrong, there is a one-sided division algorithm in $R$. So if we consider the left ideal $R(X-I)$ then the quotient space $R/R(X-I)$ is a 1-dimensional space over $\mathbb{H}$, because all the cosets of $R(X-I)$ contain a constant polynomial (divide any polynomial from the right by $X-I$ and take the remainder). Hence the left ideal $R(X-I)$ is maximal, because it is of 'codimension one'.
However, $R(X-I)$ is not an ideal of $R$. We have
$$
J(X-I)-(X-I)J=(JX-JI)-(JX-IJ)=-JI+IJ=2K.
$$
So any ideal containing $(X-I)$ must also contain the unit $2K$, and hence is equal to all of $R$.
We need to be careful here. Because $\mathbb{H}$ is not commutative, we don't get a ring homomorphism from $\mathbb{H}[X]$ to $\mathbb{H}$ by evaluating polynomials at a selected element $q\in\mathbb{H}$.
