I would try this approach. It uses, however, localizations, which is (if I remember correctly) the only tool needed for proving that the nilradical is the intersection of all prime ideals.
Let $a \in R$ be some non-invertible element. Thus, it is contained in the only maximal (= the only prime) ideal of $R$. Consider the localization $S^{-1}R$ of $R$, where $S=\{a^k \; | \; k \in \mathbb{N}\}$. Now the "correspondence theorem for localizations" says that prime ideals of $S^{-1}R$ are in one-to -one correspondence with prime ideals of $R$ not intersecting $S$. But since $a$ is a member of the only prime ideal of $R$, it follows that $S^{-1}R$ must be the zero ring (it is an unital ring with no prime ideals, hence no maximal ideals).
Thus, we have $(1/1)=(0/1)$ in $S^{-1}R$, which by definition means that $a^n=a^n(1-0)=0$ in $R$ for some $n \in \mathbb{N}$.
Another approach (more elementary one):
I have a feeling this is just a rephrasing of the proof above, however, it may be more transparent.
Let $a$ be a non-invertible element which is not nilpotent. Then $a$ is contained in some maximal ideal $M$, which is prime.
Consider the family of ideals
$$\mathcal{M}=\{I \;|\; a^k \notin I\; \forall k\}$$
Ordered by inclusion. Since $a$ is not nilpotent, $0 \in \mathcal{M}$, hence the collection is nonempty. It is clearly closed under taking unions of chains of ideals. Hence it contains some maximal element $P$.
The claim is that $P$ is prime ideal. Consider two elements $x,y \in R \setminus P$. Then we have $xR+P, yR+P \supsetneq P$. Since $P$ was maximal in $\mathcal{M}$, it follows that
$$a^m=xr+p, a^n=ys+q$$
for some $n,m \in \mathbb{N},\;\; r,s \in R, \;\; p,q \in P$. Then
$$a^{n+m}=xyrs+xrq+ysp+pq,$$
where $xrq+ysp+pq \in P$. It follows that $xyrs \notin P$, since $a^{n+m}\notin P$. Thus, $xy \notin P$.
We have proved that $P$ is a prime ideal not containing $a$. In particular, $M$ and $P$ are two distinct prime ideals in $R$. Thus, assuming $R$ has only one prime ideal, all non-invertible elements must be nilpotent.
From your comment in the comments:
Then it suffices to take some prime ideal 𝑃, apply (i) and the minimal prime ideal obtained is as in (ii), right?
This will work, yes. If you know that in a ring with identity
- There exist maximal ideals; and
- maximal ideals are prime.
then you can extract a minimal prime ideal contained in that prime ideal, which would necessarily be a minimal prime in the entire ring.
However I do not understand (ii), because if this is true, that would mean that there is some prime ideal 𝑃 such that for every prime ideal 𝑄 we have 𝑃⊂𝑄.
As also discussed in the comments, it seems you are interpreting minimal as minimum (meaning "a prime ideal contained in all other prime ideals").
A minimal prime ideal is simply one that does not properly contain any other prime ideal. In your example, $(2)$ and $(3)$ are both minimal prime ideals of $\mathbb Z_6$, and no minimum prime ideal exists in the ring.
Best Answer
When reading this post I began to wonder if $Nil(R)=\bigcap\{ P\mid P\text{ prime}\}$ was equivalent to choice, but that led me to this interesting post.
If I read it correctly, the nilradical equation is not equivalent to AC!
Hope this helps!