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.
It is equivalent to show that every non-minimal ideal is maximal. If $P\subsetneq Q\subset R$ are prime ideals, we can view $Q$ as an ideal of the integral domain $R/P$. The additive group of $R/P$ is also finitely generated, so it suffices to show that if $R$ is an integral domain with finitely generated additive group, then every non-zero prime ideal of $R$ is maximal.
Suppose $R$ is an integral domain whose additive group is finitely generated, and let $n$ be the rank of the additive group. Suppose $P\subset R$ is a non-zero prime ideal, and take $x\in P\backslash\{0\}$. Since multiplication by $x$ is a bijection of $R$ onto $xR$, the rank of $xR$ is $n$, so the rank of $P$ is also $n$. The quotient of a finitely generated abelian group by a subgroup with the same rank is finite, so $R/P$ is a finite integral domain, hence a field. It follows that $P$ is maximal.
Best Answer
Maybe no one would care about this, but I just want to write a proof about this statement. Let $R$ be a commutative ring (with $1$) and $S$ be a multiplicatively subset of $R$ (we can assume that $0\notin S$).
If we consider the natural homomorphism $\phi \colon R\rightarrow S^{-1}R$ given by $a\mapsto \frac{a}{1}$, then for ideals $I$ of $R$ and $\mathscr{I}$ of $S^{-1}R$, $I^e=\langle \phi(I)\rangle$ is an ideal of $S^{-1}R$ called the extension of $I$, and $\mathscr{I}^c=\phi^{-1}(\mathscr{I})$ is an ideal of $R$ called the contraction of $\mathscr{I}$.
Proposition.- Let the situation be as in the previous paragraph. Then there exits a bijective map $\Phi$ between the sets $A=\{P\in \text{Spec}(R): P\cap S=\emptyset\}$ and $B=\text{Spec}(S^{-1}R)$ given by $$\Phi\colon A\rightarrow B$$ $$\;\;\;\;\;P\mapsto P^e.$$
Whose inverse is $\Psi\colon B\rightarrow A$ given by $\mathscr{P}\mapsto \mathscr{P}^c$. Moreover, both $\Phi$ and $\Psi$ preserve inclusions.
Proof: This is the corollary of theorem 5.32 given in Sharp's "Steps in Commutative Algebra".
Now, in the particular case $S=R\setminus P$, where $P$ is minimal prime ideal of $R$, we have for $P'\in \text{Spec}(R)$, $P'\cap (R\setminus P)=\emptyset$ if and only if $P'\subseteq P$, which implies that $P'=P$. So it follows that $P^e=PR_P$ is the only prime ideal of $R_P$. But it's a standar result that $\text{Nil}(R_P)$ is the intersection of the elements of $\text{Spec}(R_P)$ and therefore $\text{Nil}(R_P)=PR_P$. Hence, every element of $PR_P$ is nilpotent.