Vakil's 4.5.H reads as follows
Suppose $I$ is any homogeneous ideal of $S_•$ contained in $S_+$, and $f$ is a homogeneous element of positive degree. Show that $f$ vanishes on $V(I)$ (i.e., $V(I) ⊂ V(f)$) if and only if $f^n ∈ I$ for some $n$. (Hint: Mimic the affine case; see Exercise 3.4.J.)
It seems like one cannot exactly mimic the affine case without knowing something like "for a homogenus ideal $I$, $\sqrt{I}$ is the intersection of all homogeneous primes that contain it." (EDIT: Maybe a better way of saying this is that minimal primes of a homogeneous ideal are homogeneous. )But this statement has not been mentioned so far in Vakil, so I am wondering if there is an easier way to do it.
A similar question has been asked here, but that seemed to use a lot of results that had not been developed in Vakil (yet).
$\DeclareMathOperator{\Proj}{Proj}\newcommand{\fp}{\mathfrak{p}}$Let $I$ and $f$ be as mentioned in the post.
First, assume that $V(I) \subset V(f)$. We wish to show that $f \in \sqrt{I}$.
Suppose not. Then, $f \notin \fp$ for some ordinary prime $\fp$. Now, consider its homogenisation $\fp^{\ast}$. Since $\fp^{\ast}$ is a subset of $\fp$, it follows that $f \notin \fp^{\ast}$.
On the other hand, since $I$ is homogeneous, it follows that $I \subset \fp^{\ast}$. Thus, $\fp^{\ast} \in V(I) \setminus V(f)$, a contradiction.
Conversely, assume that $f \in \sqrt{I}$. Let $\fp$ be a homogeneous prime containing $I$. Then, $f \in \sqrt{I} \subset \sqrt{\fp} = \fp$ showing that $\fp$ contains $f$. This proves $V(I) \subset V(f)$.
The only "extra bit" which looks like Vakil had not mentioned is the boldened part. That is not particularly difficult to show.
By definition, $\fp^{\ast} = \bigoplus_{n \geqslant 0} (\fp \cap S_n)$. It is clear that $\mathfrak{p}^\ast$ is an ideal and is homogeneous, almost by construction. Since $\fp^{\ast} \subset \fp$, it also follows that $\fp^{\ast}$ does not contain $S_+$.
To see that it is prime, it suffices to only check for homogeneous elements. (Exercise 4.5.C. (c))
Suppose $a, b$ are homogeneous such that $ab \in \mathfrak{p}^\ast$. Then, $ab \in \mathfrak{p} \cap S_n$, where $n = \deg(ab).$ Thus, $ab \in \mathfrak{p}$ and hence, $a \in \mathfrak{p}$ or $b \in \mathfrak{p}.$ In either case, $a \in \mathfrak{p}^\ast$ or $b \in \mathfrak{p}^\ast.$