I'm trying to prove the right ideal $(x-i)\mathbb{H}[x]$ of $\mathbb{H}[x]$ is maximal.
I've tried defining a surjective function $f:\mathbb{H}[x] \to \mathbb{H}$ by $f(x) \mapsto f(i)$ and using the 1st isomorphism theorem, but I don't think that $\ker f = (x-i)\mathbb{H}[x]$ as $f(j(x-i)) = 0$, but $j(x-i) = (x-i)j+2k \notin (x-i)\mathbb{H}[x]$ so I don't think I can use this method. Does anyone have any suggestions?
As you've discovered, evaluation no longer works like you want it to. Another illustration is that $ix=xi$, but it isn't valid if j is substituted for $x$. The problem is that $x$ is assumed to be central whereas what you substitute may not be.
However, you can, with care, still perform a one-sided division algorithm. Using it, you can write $p(X)=(x-i)q(x)+ r$ for some remainder $r\in\Bbb H$.
The map you want is the one that maps each polynomial to its remainder in this setup. The kernel is exaclty your right ideal, then, and it is obvious that this map is onto $\Bbb H$, so the kernel is maximal.
For commutative fields evaluation gives you that remainder for free, but it does not work that way anymore for division rings.