= Solution
For a multiplication operator $A_j(x)$, $[p_i,A_j]=-i\hbar\partial_iA_j$. Therefore the <mechanical momentum> satisfies
$$
[\pi_x,\pi_y]=iq\hbar(\partial_xA_y-\partial_yA_x)=iq\hbar B.
$$
The displayed ladder operators implicitly require $qB>0$, so that their denominator is real and the second is the adjoint of the first. Then $[a,a^\dagger]=1$ and
$$
H=\frac{q\hbar B}{m}(a^\dagger a+\tfrac12)+\frac{p_z^2}{2m},\qquad
\boxed{E_{n,k_z}=\frac{q\hbar B}{m}(n+\tfrac12)+\frac{\hbar^2k_z^2}{2m},\quad n=0,1,\ldots.}
$$
The free longitudinal motion gives a <continuous spectrum> in $k_z$, and the transverse <Landau levels> have guiding-centre <degeneracy of a Landau level>. For arbitrary sign of $qB$, use $|qB|$ in the <energy> and $a=(\pi_x+i\operatorname{sgn}(qB)\pi_y)/\sqrt{2\hbar|qB|}$. The question's normalizable <wavefunctions> likewise assume $qB>0$.
For the stated symmetric-gauge <wavefunction>, $|\psi|^2\propto r^{2M}e^{-qBr^2/(2\hbar)}$, whose peak for $M>0$ is at $r_M^2=2\hbar M/(qB)$. Thus the largest useful angular index is about $qBR^2/(2\hbar)$, giving
$$
\boxed{N_{\rm LLL}\simeq\frac{qBR^2}{2\hbar}=\frac{\Phi}{h/q},\qquad \Phi=B\pi R^2.}
$$
The $M=0$ state should also be included in the usual <lowest Landau level>; its omission by “positive <integer>” changes only an order-one term. Including the radial measure shifts the most probable radius to $r^2=(2M+1)\hbar/(qB)$, again only an order-one counting correction. Boundary conditions and the tails of the <wavefunctions> make this a large-flux estimate, not an exact finite-<disc> count.
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