Show that the closed points of $\operatorname{Spec} R$ are exactly the elements of $\operatorname{maxSpec} R$ that is, every closed point of $\operatorname{Spec} R$ is maximal.
Proof attempt:
Let $\{\mathfrak{p}\} \subset \operatorname{Spec}$ be closed. Then $\overline{\{\mathfrak{p}\}} = \{\mathfrak{p}\}$, but $\overline{\{\mathfrak{p}\}}$ is the smallest closed set containing $\mathfrak{p}$ and since the closed sets of the spectrum are of form $V(I)=\{ \mathfrak{p} \in \operatorname{Spec}R \mid \mathfrak{p} \supset I\}$ the smallest one containing $\mathfrak{p}$ is $V(\mathfrak{p})$, thus $$\{\mathfrak{p}\} = \overline{\{\mathfrak{p}\}} = V(\mathfrak{p})$$ and so there doesn't exist any prime ideal $\mathfrak{q}$ of $\operatorname{Spec}$ properly containg $\mathfrak{p}$. Since every maximal ideal is prime this implies that no maximal ideal contains $\mathfrak{p}$ so $\mathfrak{p}$ must be maximal.
Is the idea here correct? I'm wondering if there exists a maximal ideal that isn't a prime which would contain $\mathfrak{p}$ which would make this conclusion false.
Your argument is correct. Remember that an ideal is prime if and only if the quotient is an integral domain, and maximal if and only if the quotient is a field. Thus any maximal ideal is prime.