In standard quantum mechanics, where Hamiltonians are (selfadjoint extensions of) operators of the form $$H= -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+ V(x),$$ boundedness from above is not possible for every choice of $V$.
Here is a sketch of proof.
Boundedness from above implies that $$\langle \psi| H \psi \rangle < c$$ for some finite constant $c$ and every normalized vector $\psi$ in the domain of the (selfadjoint extension of) operator. Consider a smooth compactly supported function $\psi$ with unit norm (it obviously belongs to the domain of $H$) and define $\psi_k(x)= e^{ikx}\psi(x)$. This second function is still smooth, normalized and belongs to the domain of $H$. Integrating by parts we have that $$\langle \psi_k|H\psi_k\rangle =\frac{\hbar}{2m} \int_{\mathbb{R}} |k\psi(x)+ \psi'(x)|^2 dx + \int_{\mathbb{R}} V(x) |\psi(x)|^2dx.$$ When $k\to +\infty$, the first integral diverges to $+\infty$ whereas the latter remains finite (it does not depend on $k$). This proves that the above $c$ cannot exist.
The result easily generalizes to many dimensions.
In summary, boundedness from above is not permitted by the standard form of Hamiltonians of QM. Though it is mathematically permitted, it is not physically motivated in standard QM because it requires Hamiltonian operators if non-standard type.