Updated answer: For $k=2$ the conditions are contradictory.
We have the decomposition $H=H_0\oplus H_1$. You now impose the following conditions: (1) $TH_0=H_1$; (2) $T$ injective on $H_0$ and $H_1$; (3) $T^2 H_0=H_0$; (4) $Tu=0$ for some $u\notin H_1$.
Since $N(T^2)=N(T)$ for the self-adjoint operator $T$, we can rephrase the last condition as: (4') $T^2u=0$ for some $u\notin H_1$.
Let's write $$ T= \begin{pmatrix} 0 & B \\ B^* & C \end{pmatrix} , $$ and here the components refer to this decomposition of $H$; the zero in the $(1,1)$ is a consequence of (1) above.
In terms of $B,C$, your conditions become: (1') $B^*$ surjective (onto $H_1$); (2') $B^*$ injective (on $H_0$) and also if $By=Cy=0$, then $y=0$; (3') $BC=0$, $BB^*$ surjective; (4'') $BB^*x=$ for some $x\in H_0$, $x\not= 0$.
Since $B^*$ is injective, (4'') shows that $N(B)\not= 0$, but since $N(B)=R(B^*)^{\perp}$, this contradicts (1').
Comment: The part below answers the original version of the question.
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.