Nilpotent elements of $R$: If $x \in \operatorname{Nil}(R)$ then $1+x \notin P$, where $P$ is any prime ideal of $R$. Denote $\operatorname{Nil}(R)$ as $N$. We already have than $N$ is an ideal in $R$. It remains to prove that:

$N \subseteq P$ for any prime ideal, $P$,  in $R$ and then conclude that $\forall x \in N, 1+x \notin P$

Let $x \in N$. $\exists r \mid x^r =0 \in P$. By induction suppose $r = 2$: $x^2 \in P \implies x \in P$. Induction hypothesis: $x^k \in P \implies x \in P$: $x^{k+1} \in P \implies x \in P $ or $x^k \in P$. Hence $N\subseteq P$.
So this next part is where I'm feeling less confident, because it seems too easy...
$x \in N $ gives $-x \in N$ so $1+x \in N \implies$ $1 \in N$, a contradiction since $\nexists r \in \mathbb{N} | 1^r =0$? And therefore $1+x \notin N \implies 1+x \notin P$ ?
edit: fixing the second part:
suppose $x \in N$, and $1+x \in P$. Then $N \subseteq P$ gives  $-x \in P \implies 1 \in P$, but a prime ideal cannot equal the whole ring.
edit2: Thank you all!
 A: The first part of your argument is correct, but is not formulated very clearly. More words and fewer symbols, as well as more spacing, would help a lot. For example:

Let $x\in N$. Then $x^r=0$ for some positive integer $r$, and so $x^r\in P$. Then also $x\in P$ by induction:
Induction hypothesis: If $r=2$ then $x^2\in P$ and so $x\in P$.
Induction step: If $r>2$ then $x^r\in P$ and so either $x$ or $x^{r-1}\in P$. Hence by induction, if $x^r\in P$ for some positive integer $r$, then $x\in P$.

You could even take $r=1$ for your induction hypothesis. You could also argue slightly less formally:

If $x^r\in P$ then either $x\in P$ or $x^{r-1}\in P$. Repeating this argument $r-1$ times shows that $x\in P$.

Your proof of the second part is incorrect. It would also be helped by some more words and fewer symbols. Your argument shows that $1+x\notin N$. This does not imply that $1+x\notin P$ for every prime ideal $P$. After all $N$ is just a subset of every prime ideal $P$; every $P$ may contain elements that are not contained in $N$, for example $1+x$.
Instead start by supposing toward a contradiction that $1+x\in P$ for some prime ideal $P$ and some $x\in N$. Then also $-x\in N$ and hence $-x\in P$. It follows that $1\in P$, a contradiction.
A: The first part looks good while the second needs some more work.
The first part could be written clearer using that $x^{n+1}=x\cdot x^n$ to deduce that if a power of an element $x$ is contained in a prime ideal $P$ then the element itself is in $P$. Your argument uses this idea but is written a bit more complicated than necessary.
For the second part let us introduce some more quantifiers. If $x\in N$ then $x\in P$ for all prime ideals $P$. What you show is that there is some prime ideal $P$ such that $1+x\notin P$ as otherwise $1\in N$ giving your contradiction. What you want to show, however, is that $1+x\notin P$ for all prime ideals $P$.
Well, note the following:
$$
\frac{1-(-x)^n}{1-(-x)}=\sum_{i=0}^{n-1}(-1)^ix^i \iff 1-(-x)^n=(1+x)\sum_{i=0}^{n-1}(-1)^ix^i
$$
Now suppose that $x$ is nilpotent (i.e. $x\in N$). What can you deduce from this identity then? Moreover, what does this tell us about containment in prime ideals?

There is an alternative for the second part: Consider the intersection of all maximal ideals and call it the Jacobson radical $J=\operatorname{Jac}(R)$. Note that $N\subseteq J$ as all maximal ideals are prime. Now show that $x\in J$ iff $1+xy$ is a unit for all $y\in R$ (I assume that all rings are commutative and unital). This statement is more general and crucially relies on the fact that every proper ideal is contained in some maximal ideal.
A: Your proof for the implication $N \subset P$ is correct.
Note however, that $N \subset P$ does not give $1+x \notin N \implies 1+x \notin P$. That would be the implication $N^c \subset P^c$, where ${}^c$ denotes the complement. This is set-theoretically wrong.
However, if you replace $N$ by $P$, you get almost the correct answer:
$x \in N$ gives you $x \in P$ by the first part, and so also $-x \in P$. Then $1+x \in P$ would imply $1 = 1+x - x \in P$, which means $P = R$ (by the multiplicative property of an ideal). But a prime ideal is never the full ring, contradiction.
