In the $2\times 2$ matrices over $\mathbb{Z}$ (or over $\mathbb{Q}$, or over $\mathbb{R}$), the elements
$$x = \left(\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right)\quad\text{and}\quad y = \left(\begin{array}{cc}
0 & 0\\
1 & 0
\end{array}\right)$$
are nilpotent: $x^2=y^2 = 0$. But
$$x+y = \left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right)$$
is not only not nilpotent, it's a unit: $(x+y)^2 = 1$. So the set of nilpotent elements is not a subgroup, hence cannot be an ideal.
In noncommutative rings, you want to consider instead the notion of "nilpotent (left/right/two-sided) ideal": an ideal $I$ for which there exists $k\gt 0$ such that $I^k=0$. The sum of any finite family of nilpotent (left/right/two-sided) ideals is again a nilpotent (left/right/two-sided) ideal. An alternative is to consider "nil ideals", which are ideals in which every element is nilpotent. This is generally the case: when going from the commutative setting to the non-commutative setting, one often switches from "element-wise" to "ideal-wise" conditions. Thus, an ideal $P$ in a non-commutative ring is a prime ideal if for any two ideals $I$ and $J$, if $IJ\subseteq P$ then either $I\subseteq P$ or $J\subseteq P$ (compare to the definition of prime ideal in a commutative setting, which is called a "totally prime" or "completely prime" ideal in the noncommutative setting).
(Assuming the rings are commutative with identity, else the statement is false.)
As Norbert says, your two useful facts immediately combine to say that the set of nilpotent elements is equal to the intersection of prime ideals containing the ideal $\{0\}$, and that means all prime ideals.
First, I need to show that the nilpotent elements are in fact an ideal
Well sure, this is possible, although it is not necessary. You can find how to do this in a few other questions on the site, starting here
The set of all nilpotent elements is an ideal
and then I need to show it's prime.
That... is not in the question at all. Showing something is equal to an intersection of prime ideals does not mean it itself is prime. Most of the time, the ideal of nilpotent elements is not prime.
I just don't know how to make this proof flow and exactly what leads to what.
Ok, fair enough. The outline of the proof is: show the two sets are equal by showing each one is a subset of the other.
On one hand, each nilpotent element is contained in each prime ideal. Thus the nilpotent elements are in the intersection of primes.
The other direction is a little harder, and usually we use a lemma: every ideal maximal with respect to being disjoint from a multiplicative subset of R is a prime ideal of R. Using this, we can show that for each nonnilpotent element, there is a prime ideal disjoint from the powers of that element, and hence the prime does not contain the element. This shows how nonnilpotents are excluded from the intersection, so that everything there is nilpotent.
As a corollary, you get that the nilpotents form an ideal, and it is not really necessary to prove this ahead of time.
Best Answer
Recall that $\frac{x}{1}=0$ implies that there is an $s\in S$ such that $xs=0$ (for example look at page 37 of Atiyah-MacDonald). Since $\operatorname{Ann(x)}$ is an ideal and $x$ is nilpotent, then there is a prime ideal $\mathfrak{p}$ such that $x^{n-1}\in\operatorname{Ann}(x)\subseteq\mathfrak{p}$ and $x\in\mathfrak{p}$. Thus by contraposition $\frac{x}{1}\neq 0$ in $R_{\mathfrak{p}}$.
Your proof is essentially a full write-down of the nilradical proof:
By Corollary 3.12 if $\mathfrak{N}$ is the nilradical of $R$, then $\mathfrak{N}_\mathfrak{p}$ is the nilradical of $R_\mathfrak{p}$. Since being the $0$ module is a local property, $\mathfrak{N}_\mathfrak{p}=0$ for every prime ideal $\mathfrak{p}\subset R$ implies that $\mathfrak{N}=0$.