= Solution
The Hamiltonian is the sum of three commuting one-dimensional <quantum harmonic oscillator>[harmonic-oscillator] Hamiltonians. Its product <eigenstate>[eigenstates] are
$$
\Psi_{n_xn_yn_z}(x,y,z)
=\psi_{n_x}(x)\psi_{n_y}(y)\psi_{n_z}(z),
$$
with energies
$$
\boxed{
E_{n_xn_yn_z}
=\hbar\omega\left(n_x+n_y+n_z+\frac32\right)
}.
$$
Equivalently, the level with $N=n_x+n_y+n_z$ has energy $E_N=\hbar\omega(N+3/2)$, as in the <three-dimensional isotropic harmonic oscillator>.
Up to normalization, the ground-state wavefunction is
$$
\Psi_{000}
=\exp\left[-\frac{M\omega}{2\hbar}(x^2+y^2+z^2)\right].
$$
It is radial, so $\nabla\Psi_{000}$ is parallel to the position vector. Since the <orbital angular momentum> is $L=-i\hbar\,r\times\nabla$, every component of $L$ annihilates this state. Therefore
$$
L_z\Psi_{000}=0,
\qquad
L^2\Psi_{000}=0,
$$
and the ground state has $\ell=m=0$.
The first excited level has $N=1$ and is spanned by
$$
x\Psi_{000},\qquad y\Psi_{000},\qquad z\Psi_{000}.
$$
The state
$$
\Phi_0=z\Psi_{000}
$$
satisfies $L_z\Phi_0=0$. Applying the ladder operators gives
$$
L_+\Phi_0=-\hbar(x+iy)\Psi_{000},
\qquad
L_-\Phi_0=\hbar(x-iy)\Psi_{000}.
$$
Hence convenient $m=\pm1$ eigenstates are
$$
\boxed{\Phi_{\pm1}=(x\pm iy)\Psi_{000}},
$$
up to normalization and irrelevant overall phases.
Finally, the isotropic Hamiltonian is rotationally invariant. The <rotational invariance of a central-potential Hamiltonian> gives
$$
[H,L_z]=[H,L^2]=[L_z,L^2]=0.
$$
These commuting self-adjoint operators can be simultaneously diagonalized within each energy eigenspace, which is why joint eigenstates of $H,L^2,L_z$ must exist.
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