Any two invariant two dimensional subspaces must intersect in an eigenvector. If such a subspace contains $e=(0\ 0\ 1)^t$, suppose $v$ is a vector in the subspace that is not a multiple of $e$. The only possibility is for the second component to be $0$, otherwise $v,Av, e$ is a basis for the whole space. Thus there is only one invariant two dimensional subspace containing $e$, and it contains the other eigenvector. Call this subspace $U$.
If $U'$ is another invariant two dimensional subspace it follows that $U\cap U'$ is the one dimensional subspace spanned by $(1\ 0\ 0)^t$. Let $v'=(x\ y\ z) ^t$ be a vector completing a basis for $U'$. Then necessarily $y\neq 0$. If $z\neq 0$, then $U'$ contains $e$ as well, which is impossible. Thus the second invariant subspace, the only other one, is the one corresponding to the Jordan block with eigenvalue $2$.
I'm guessing these are the two subspaces you found.