Give an example of a Hilbert space $\mathcal{H}$ and an operator $A: \mathcal{H} \to \mathcal{H}$ satisfying $$\sigma \left( A \right) = \varnothing \neq \sigma \left( A^2 \right),$$ where $\sigma \left( A \right)$ denotes the spectrum of $A$.
Can't really get anywhere with this. I know that, by the fundamental theorem of algebra, $\mathcal{H}$ has to be infinite-dimensional. Then what comes to mind is looking at $\mathcal{l}^2 \left( \mathbb{R} \right)$, but that's it.
On
It's impossible to find such an operator. Suppose that $\lambda$ is in the spectrum of $A$, then $$ (A^2 - \lambda I) = (A - \sqrt \lambda I)(A + \sqrt \lambda I) $$ $(A^2 - \lambda I)$ fails to be bijective, which means that one of the operators $A \pm \sqrt \lambda I$ must not be bijective, which means that the spectrum of $A$ cannot be empty.
Since you spoke about $\ell^2(\mathbb R)$, I deduce that your space doesn't have to be complex.
Hence, you may take $$ A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. $$
This matrix has no real eigenvalues, since $$ \det \begin{pmatrix} -\lambda & 1 \\ -1 & -\lambda \end{pmatrix} = \lambda^2 + 1, $$ but $$ A^2 =\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, $$ which has the eigenvalue $-1$.