Closed subset $Z \subseteq \operatorname{Spec} A$ This is an argument in Gortz algebraic Geometry I text book pg85. I extracted what I don't understand.
He proves the following

Let $A$ be a unital commutative ring. $Z \subseteq  \operatorname{Spec} A$ a closed subset satisfying the following property: Given $s \in A$,
If $Z \subseteq V(s)$, then $s^n=0$ for some $n$.

He concludes

$V(s)=X$ and $Z=X$.

I don't understand this argument. What facts did we used here?
EDIT: The following is my proposed proof: $V(s)=X$. Since if $s$ is nilpotent, it lies in the nil radical, hence intersection of all primes.
Now $Z=V(I) = V(\sum_{i \in I} A i) = \bigcap_{i \in I} V(i) = X$.
I have the good idea. I would emphasize on the fact that $V(i)=X$ since $i$ is nilpotent by hypothesis.

Since $Z$ is closed, $Z=V(I)$, this implies that for every $s\in I$, $Z\subset V(I)\subset V(s)$, we deduce that $s$ is nilpotent and $I\subset nil(A)$, $X=V(nil(A))\subset V(I)$ implies that $V(I)=Z=X$.
I have the good idea. I would emphasize on the fact that $V(i)=X$ since $i$ is nilpotent by hypothesis.

Since $Z$ is closed, $Z=V(I)$, this implies that for every $s\in I$, $Z\subset V(I)\subset V(s)$, we deduce that $s$ is nilpotent and $I\subset nil(A)$, $X=V(nil(A))\subset V(I)$ implies that $V(I)=Z=X$.