The eigenspaces of $A$ are the kernels of $A-1$ and $A-2$. Since the multiplicity of eigenvalue 2 is one, it has no generalized eigenvalues.
For the eigenvectors with eigenvalue $1$, we find the kernel of $A-1$.
We want to solve
0 & 0 & 0\\
x_1 & 2 & 0\\
x_2 & x_3 & 1
So we get $x_1v_1+v_2=0$ and $x_2v_1+x_3v_2=0,$ and substitution gives $(x_2-x_1x_3)v_1=0.$
So I agree that $(0,0,1)$ is the only independent eigenvector in this kernel, but only if we assume $x_2-x_1x_3\neq 0$, so that we can solve the system with $v_1=v_2=0$.
On the other hand, if $x_2-x_1x_3=0$, then $(1,-x_1,0)$ and $(0,0,1)$ are both eigenvectors. In this case the matrix is diagonalizable, which also tells you its Jordan form.
So assuming that $x_2-x_1x_3\neq 0$, to find the generalized eigenvectors for eigenvalue 1, find the kernel of $(A-1)^2$. There is one independent vector there that is not also in the kernel of $A-1$ (so not an eigenvector), which is $(1, -x_1, 0)$.
In general to find the generalized eigenvectors for an eigenvalue $c$, you compute the kernel $(A-c)^k$, where $k$ is the power with which the $(x-c)$ factor occurs in the minimal polynomial.
Finally once you have the generalized eigenvector, you can write the Jordan canonical form by expressing the matrix in the ordered basis $(0,0,x_2-x_1x_3), (1,-x_1,0),(0,1,x_3)$. You get
1 & 1 & 0\\
0 & 1 & 0\\
0 & 0 & 2