Jordan form exercise What am I doing wrong?
I've been learning how to put matrices into Jordan canonical form and it was going fine until I encountered this $4 \times 4$ matrix:
$A=\begin{bmatrix} 
2 & 2 & 0 & -1 \\ 
0 & 0 & 0 & 1 \\
1 & 5 & 2 & -1 \\
0 & -4 & 0 & 4 \\
\end{bmatrix}
$
Which has as only eigenvalue $\lambda_1=\lambda_2=\lambda_3=\lambda_4=2$ with 2 corresponding eigenvectors, which I will for now call $v_1$ and $v_2$:
$v_1 = \pmatrix{0\\0\\1\\0}, v_2=\pmatrix{-3 \\ 1 \\ 0 \\ 2}
$ 
2 eigenvectors means 2 Jordan blocks so I have 2 possibilities:
$J= \pmatrix{2 & 1 & 0 & 0 \\ 
0 & 2 & 1 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2} $ or $ J= \pmatrix{2 & 1 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 0 & 2}
$
I consider the first possibility. This gives me the relations:
$Ax_1=2x_1 \\
Ax_2=x_1+2x_2 \\
Ax_3=2x_3+x_2 \\
Ax_4=2x_4 \\
$
where $x_1$ and $x_4$ should be $v_1$ and $v_2$. From the second relation $(A-2I)x_2=x_1$ I see
$\pmatrix{0 & 2 & 0 & -1 \\
0 & -2 & 0 & 1 \\
1 & 5 & 0 & -1 \\
0 & -4 & 0 & 2} \pmatrix{a \\ b \\ c \\ d} =\pmatrix{0 \\ 0 \\ 1 \\ 0} $ 
( $v_2= \pmatrix{ -3 \\ 1 \\ 0 \\ 2}
$ will give an inconsistent system)
Now I get that $x_2 = \pmatrix{-2 \\ 1 \\ 0 \\ 2}
$
From the third relation $(A-2I)x_3=x_2$:
$\pmatrix{0 & 2 & 0 & -1 \\
0 & -2 & 0 & 1 \\
1 & 5 & 0 & -1 \\
0 & -4 & 0 & 2} \pmatrix{e \\ f \\ g \\ h} =\pmatrix{-2 \\ 1 \\ 0 \\ 2} $ 
But this system is inconsistent as well! No matter which vectors I try in which places, when I try to generalize eigenvectors I seem to always end up with some inconsistency.
Is there something staring me in the face that I am overlooking? Or am I doing it completely wrong (even though this method worked fine for me before)?
Sorry for the lengthiness and thank you in advance.

Take a look at
$$
       (A-2I)^2 = \begin{bmatrix} 
0 & 2 & 0 & -1 \\ 
0 & -2 & 0 & 1 \\
1 & 5 & 0 & -1 \\
0 & -4 & 0 & 2 \\
\end{bmatrix}^2 
$$
The $a_{1,2}$ entry is non-zero. What does that tell you?
