All nilpotent $2\times 2$ matrices

Suppose that $A$ ($2\times 2$) is nilpotent. Then $\det(A)$ is $0$, implying that an eigenvalue of $A$ is real and $0$. Because $A$ is $2\times 2$, there is one other eigenvalue which must also be $0$. (If $A^n=0$, then the eigenvalues of $A^n$ are $0$ but these are just the eigenvalues of $A$ raised to the same power $n$.) At this point, you can infer that $\text{tr}(A)=0$ and $A$ itself must necessarily take the form $$ A=\begin{pmatrix}x & a \\ b & -x\end{pmatrix},\quad x\in\mathbb{R},\quad ab=-x^2. \tag{$*$} $$ It turns out that ($*$) is also sufficient for nilpotency: $$ \begin{pmatrix}x & a \\ b & -x\end{pmatrix}\cdot \begin{pmatrix}x & a \\ b & -x\end{pmatrix}=\begin{pmatrix}x^2+ab & xa-ax \\ bx-xb & ab+x^2\end{pmatrix}=\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}\cdot $$ Thus $A$ is nilpotent iff $A$ takes the form given in ($*$).

My mistake... in my comment above, I should not have said to solve the equations.

Rather: if you pick an arbitrary $P$, and compute $P^{-1} J P$, you get a nilpotent matrix. Using

$$ P = \begin{bmatrix} a & b \\ c & d\end{bmatrix} $$ but writing $P J P^{-1}$, I got $$ PJP^{-1} = D \cdot \begin{bmatrix} -ac & a^2 \\ -c^2 & ac\end{bmatrix} $$ where $D$ is the determinant $(ad - bc)$.

Such matrices (for $a \ne 0$) have the general form

$$ A = \begin{bmatrix} -S & T \\ -\frac{S^2}{T} & S\end{bmatrix}. $$ and this is a "parameterization" (with parameters $S$ and $T$) of almost all possible nilpotent matrices. We also need to add in the $a = 0$ case, i.e. $$ A = \begin{bmatrix} 0 & 0 \\ c & 0\end{bmatrix}. $$

Note that in this parameterization, it's essential that $T \ne 0$. Because of this, you can say that up to scalar multiples, all nilpotent matrices have the form $$ A = \begin{bmatrix} -S & 1 \\ -S^2 & S\end{bmatrix}. $$ or $$ A = \begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix}. $$