Let us assume $\Omega$ is sufficient smooth. Let $H=L^2(\Omega)$ and define $A:D(A)\subset H\to H$ by $Au=\Delta u$, with $D(A)=H_0^1(\Omega)\cap H^2(\Omega).$
Brezis's book, Brezis, Functional Analysis... page 327, says:
It follows from Theorem 9.25 that $D(A^\ell)\subset H^{2\ell}(\Omega)$, for every, with continiuous injection.
Here is the Theorem 9.25:
Let $\Omega$ be an open set of class $C^2$ with $\Gamma$ bounded. Let $f\in L^2(\Omega) $ and let $u \in H_0^1(\Omega)$ satisfy $$\int_\Omega \nabla u\nabla \varphi +\int_\Omega u\varphi=\int_\omega f\varphi,\ \ \forall \varphi\in H_0^1(\Omega). \tag{48}\label{48}$$ Then $u\in H^2(\Omega)$ and $\|u\|_{H^2}\leq C\|f\|_{L^2}$, where $C$ is a constant depending only on $\Omega$. Furthermore, if $\Omega$ is of class $C^{m+2}$ and $f\in H^m(\Omega)$, then $u\in H^{m+2}$ and $\|u\|_{H^{m+2}}\leq C\|f\|_{H^m}$.
I was not able to prove that $D(A^\ell)\subset H^{2\ell}(\Omega)$, even for $\ell=2$. What I have tried was the following:
By definition we have that $$D(A^2)=\{u\in H_0^1(\Omega)\cap H^2(\Omega);\ \Delta u\in H_0^1(\Omega)\cap H^2(\Omega)\}.$$ I Know that \eqref{48} means $u$ satisfies $$-\Delta u+u=f\tag{*}\label{*}$$ in the sense of distribution. Hence, I defined $v=u_{x_ix_j}$ and, since $\Delta u\in H^2(\Omega)$, I got that $$-\Delta v+v=(\Delta u)_{x_ix_j}+u_{x_ix_j}:=f\in L^2(\Omega).$$ That is, $v$ satisfies \eqref{*} and therefore satisfies \eqref{48}. But I don't know if $v\in H_0^1(\Omega)$, so I can't apply Theorem 9.25 in order to conclude that $v\in H^2(\Omega).$
It turns out that the answer is obvious, but I couldn't see it and a professor of mine show it to me. I will post the answer in case someone has the same question.
Given $u\in D(A^2)$, we define $f:=u-\Delta u$. Hence $f\in H^2(\Omega)$ and $u\in H_0^1(\Omega)$ satisfies (48). Therefore, from Theorem 9.25, $u\in H^4(\Omega)$.