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).
I think I have one. Let $k$ be the field with $2$ elements. Let $R$ be the $k$-algebra with generators $x$, $y$ and $z$, modulo the relations
$$zx=xz,\ zy=yz,\ yx=xyz.$$
It is not hard to see that monomials of the form $x^i y^j z^k$ are a basis for $R$.
We will call these the standard monomials.
For any $f \neq 0$ in $R$, write $f = \sum f_{ij}(z) x^i y^j$. We will define the leading term of $f$ to be the term $f_{ij}(z) x^i y^j$ where we choose $i+j$ as large as possible, breaking ties by choosing the largest possible power of $i$.
Lemma The center of $R$ is $k[z]$.
Proof Let $Z$ be central and write $Z$ in the basis of standard monomials. Since $Zx=xZ$, there are no powers of $y$ in $Z$. Since $Zy=yZ$, there are no powers of $x$ in $Z$. $\square$.
Lemma Every automorphism of $R$ acts trivially on the center of $R$.
Proof Every automorphism of $k[z]$ is of the form $z \mapsto az+b$ for $a \in k^{\ast}$. Since $k$ has two elements, we must have $\sigma: z \mapsto z+b$. Any automorphism of $R$ descends to an automorphism of the abelianization, which is $k[x,y,z]/(xy(z-1))$. Since $z-1$ is a zero divisor in the abelianization, $z+b-1$ must be a zero divisor as well, and this forces $b$ to be zero. $\square$.
Lemma If $f$ and $g$ have leading terms $f_{ij}(z) x^i y^j$ and $g_{kl}(z) x^k y^l$, then the leading term of $fg$ is $z^{jk} f_{ij}(z) g_{kl}(z) x^{i+k} y^{j+l}$.
Proof A computation. $\square$
Now, suppose that we have an automorphism $x \mapsto X$, $y \mapsto Y$, $z \mapsto z$ of $R$. Let the leading terms of $X$ and $Y$ be $f(z) x^i y^j$ and $g(z) x^k y^l$.
Lemma The vectors $(i,j)$ and $(k,l)$ are linearly independent.
Proof We are supposed to have $YX=zXY$. Taking leading terms
$$z^{il} f(z) g(z) x^{i+k} y^{j+l} = z^{jk+1} f(z) g(z) x^{i+k} y^{j+l}.$$
So $il-jk=1$ and $\det \left( \begin{smallmatrix} i & j \\ k & l \end{smallmatrix} \right)=1$. $\square$
Consider the images $z^a X^b Y^c$ of the standard monomials. Their leading terms are $z^{a+b+c} f(z)^b g(z)^c x^{bi+ck} y^{bi+cl}.$ Using the above lemma, these leading terms are all distinct. So there is no cancellation of leading terms in any sum $\sum s_{abc} z^a X^b Y^c$. So we see that every element in the image of the automorphism must have a leading term of the form $h(z) x^{bi+ck} y^{bj+cl}.$
But automorphism are surjective! So $(1,0)$ and $(0,1)$ must be positive integer combinations of $(i,j)$ and $(k,l)$. So either $(i,j) = (1,0)$ and $(k,l) = (0,1)$ or viceversa. We see that $X$ and $Y$ are of degree $1$ in $x$ and $y$. From this point, it's an easy computation.
Best Answer
Let $x\in R$ be nilpotent and $r\in R$ be arbitrary. If $xr$ is not nilpotent, it is invertible, so $xr(xr)^{-1}=1$ and therefore $x$ is right invertible. This is impossible, because from $x^n=0$ and $xy=1$ we get $x^{n-1}=0$, leading to a contradiction.
Similarly, $rx$ is nilpotent.
Suppose $x$ and $y$ are nilpotent, but $z=x+y$ is invertible. Then $$ (x+y)z^{-1}=1 $$ and so $$ xz^{-1}=1-yz^{-1} $$ By what we have proved before, we know that $yz^{-1}$ is nilpotent. However, when $a\in R$ is nilpotent, $1-a$ is invertible; indeed, assuming $a^{n+1}=0$, we have $$ (1-a)(1+a+\dots+a^{n})=1-a^{n+1}=1 $$ and similarly on the other side. Therefore $xz^{-1}$ is invertible. Contradiction.