= Solution
Put $e=E_+$ and $f=E_-$. The supplied identities give
$$
Hf^kv=(\chi-2k)f^kv,\qquad
ef^kv=k(\chi-k+1)f^{k-1}v.
$$
If no $f^kv$ vanishes, this is the first alternative. Otherwise choose the least $N\ge1$ with $f^Nv=0$. Applying $e$ gives $N(\chi-N+1)f^{N-1}v=0$. Since $f^{N-1}v\ne0$ and the field has characteristic zero, $\chi=N-1$. Thus
$$
\boxed{\chi\in\mathbb Z_{\ge0},\quad f^kv\ne0\ (0\le k\le\chi),\quad f^kv=0\ (k>\chi).}
$$
The endpoint must include zero: the trivial one-dimensional <module> has $\chi=0$ and $fv=0$, so the printed claim of strict positivity has that counterexample. All nonzero vectors in the displayed chain have distinct $H$-weights, and hence are linearly independent.
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