The nonzero nilpotents of M_2(R) form two components

Idea. The two equations $\operatorname{tr}=\det=0$ leave only a "cone" in $\mathbb{R}^4$; removing the vertex $0$ separates the cone into two nappes, distinguished by the sign of $\operatorname{Im}z$.

Step 1: nilpotent iff $\operatorname{tr}=\det=0$. If $A$ is nilpotent, all its eigenvalues vanish, so $\chi_A(t)=t^2$, i.e. $\operatorname{tr}A=\det A=0$. Conversely, if $\operatorname{tr}A=\det A=0$, Cayley-Hamilton gives $A^2=0$.

Step 2: coordinates. The two equations become $\operatorname{Re}z=0$ and $|w|^2=|z|^2$. Write $z=iq$ with $q\in\mathbb{R}$; then $|w|=|q|$. The matrix is nonzero iff $(z,w)\ne(0,0)$ iff $q\ne0$ (if $q=0$ then $z=0$ and $|w|=0$).

Step 3: structure. So
$$\mathcal N^*\cong\{(q,w):q\in\mathbb{R}\setminus\{0\},\ |w|=|q|\}.$$
The map $(q,w)\mapsto\bigl(q,\ w/|w|\bigr)\in(\mathbb{R}\setminus\{0\})\times S^1$ is continuous with continuous inverse $(q,u)\mapsto(q,|q|u)$. Thus
$$\mathcal N^*\cong(\mathbb{R}\setminus\{0\})\times S^1=\bigl((-\infty,0)\times S^1\bigr)\sqcup\bigl((0,\infty)\times S^1\bigr).$$

Step 4: components. Each piece is a product of connected spaces, hence connected. The sign of $q=\operatorname{Im}z$ is a continuous function on $\mathcal N^*$ taking the two values $\pm$, so the two pieces are open and closed in $\mathcal N^*$. Hence they are exactly the connected components.

Check. $E=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ has $a=d=c=0$, $b=1$, so $z=-\frac i2$, $w=\frac i2$: $q=-\frac12<0$ and $|w|=\frac12=|q|$. The matrix $E^T=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ has $z=\frac i2$, $w=\frac i2$: $q=+\frac12$, on the other component.

Remark. Including $0$, the set $\{A^2=0\}=\{|w|=|\operatorname{Im}z|,\ \operatorname{Re}z=0\}$ is a double cone over a circle in a 3-dimensional subspace; it is connected (through its vertex $0$), and removing the vertex disconnects it. For $n\ge3$ the situation is different: the nonzero nilpotent cone of $M_n(\mathbb{R})$ is connected (it is star-shaped about $0$ and $0$ can be approached through other orbits).

Lesson. A cone minus its vertex can be disconnected; a continuous sign function (here $\operatorname{sign}\operatorname{Im}z$) is the tool to prove disconnection.