This is what I have so far:
Let $z = 2 + i$. Then \begin{align} z - 2 &= i \\ (z - 2)^2 &= -1 \\ z^2 - 4z + 5 &= 0 . \end{align} Define $q(z) = z^2 - 4z + 5$ so that $q \in \ker(\varphi)$ and $(q) \subseteq \ker(\varphi)$. Now $q$ is irreducible over $\mathbb{R}$, so $(q)$ is a maximal ideal of $\mathbb{R}[x]$ (proof ommitted).
The tutorial solution then says something like
...therefore $\ker(\varphi) = (q)$ because $\ker(\varphi)$ is nontrivial.
I don't understand this last sentence. How can we show $\ker(\varphi) \subseteq (q)$?
We have $(q)\subset \ker(\varphi)\subsetneq\mathbb{R}[X]$ (since $\varphi$ is not identically zero)
Now $(q)$ is a maximal ideal, contained in another ideal $\ker(\varphi)$, so by maximality $(q)=\ker(\varphi)$.
(An ideal $M$ of a commutative ring with $1$ is maximal if, $M\neq R$ and for any ideal $I$ containing $M$, either $I=M$ or $I=R$)
Note that the maximality argument maybe avoided. If $f$ lies in $\ker(\varphi)$, we can divide $f$ by $q$, and get $f=pq+aX+b,$ for some $a,b\in\mathbb{R}$. Evaluating at $z$ yields $$0=f(z)=p(z)q(z)+az+b=p(z)\cdot0+az+b=az+b=2a+b+ia.$$ Hence $2a+b=0$ and $a=0$, so $a=b=0$ and $f=pq\in (q)$.