You want to find matrices of the form $A=\begin{pmatrix} x \quad y \\ 0 \quad z\end{pmatrix}$ which are similar to another matrix of the form $B=\begin{pmatrix} 0 \quad b \\ c \quad d\end{pmatrix}$.
Certainly we must have the same determinant and trace. So $x+z = d$ and $xz = -bc$. Note that in the $2 \times 2$ case, this is enough to determine the characteristic polynomial of the given matrix, since this determines the roots of the characteristic polynomial, which is of degree $2$ and would thus have roots $x,z$. I erred in saying that two matrices with the same characteristic polynomial are similar in the $2 \times 2$ case. Thus, the eigenvalues of $B$ are also precisely $x,z$.
Now, by the JCF, $A,B$ are similar to only two kinds of matrices : one which is diagonal with entries $x,z$ (this is implied by, but not equivalent to $x \neq z$) and the other which is upper triangular with the same entries $x=z$ on the diagonal, and $1$ on the top right corner. If $A,B$ have the same JCF, then they will be similar.
Now, suppose $x \neq z$, then $A,B$ will have distinct eigenvalues, implying that both of them have JCF diagonal with entries $x,z$, showing that they are similar , provided we choose $d = x+z$ and $bc$ to be any entries such that $-bc = xz$.
Suppose $x = z$ now. We note that if $A$ has eigenvector $v= (v_1,v_2)$ then $xv_1+yv_2 = xv_1$ by comparing first entries of $Av = xv$.
In similar fashion, $Bw =xw$ implies $bw_2 = xw_1$ and $cw_1 = (x-d)w_2 = -xw_2$. Note that $x = 0$ implies $A$ is already a matrix of the form that $B$ is, so we take $x \neq 0$, then note that $bw_2 = xw_1$ again implies that only one dimension is spanned by an eigenvector of $B$. Thus, $B$ has JCF as a single Jordan block.
Suppose $y \neq 0$. Then, this forces $v_2 = 0$, so we get a unique eigenvector $(1,0)$. It follows that $A$ must have JCF as a single Jordan block with $1$ on the top right. Now, the JCF of $A$ matches the JCF of $B$ so $A,B$ are similar.
But if $y = 0$, then indeed $A$ is a diagonal matrix itself, so doesn't have the same JCF as $B$ anyway, showing that $A,B$ can't be similar.
Finally, the conclusion is that $y \neq 0$ or $x \neq z$ are necessary and sufficient for $A$ to be similar to a matrix of the form $B$.