Comment: This answers the original version of the question, not the current one after an edit.
It does follow, but in a rather trivial, disappointing way. Let $S$ be the operator $T^k$, restricted to its reducing subspace $H_0$. Since $N(T^k)=N(T)$ for a self-adjoint operator, we have $N(S)= N(T)\cap H_0$. By assumption, if $x\in H_0$, $x\not= 0$, then $x\notin N(T)^{\perp}$, so $x\notin N(S)^{\perp}$. Thus $S=0$ and hence $T=0$ as well on $H_0$. But then $H=H_0$, and your operator was the zero operator all along.