Spherical co-ordinates are $(r,\theta,\phi)$. The momenta corresponding to $\theta$ and $\phi$ are $L_x$ and $L_z$.
No they aren't. It's true that $p_\phi=L_z$, but $p_\theta= \cos(\phi) L_y - \sin(\phi) L_x$.
The free-particle Lagrangian in spherical coordinates is
$$L = \frac{1}{2} m\big(\dot r^2 + r^2\dot \theta^2 + r^2\sin^2(\theta) \dot \phi^2\big)$$
$$\implies \pmatrix{p_\theta \equiv \frac{\partial L}{\partial \dot\theta} = mr^2\dot \theta\\p_\phi \equiv \frac{\partial L}{\partial \dot \phi} = mr^2\sin^2(\theta) \dot \phi}$$
Recall that
$$L_x = m(y \dot z - z \dot y) = m r^2\big(-\dot \theta \sin^2(\theta)\sin(\phi)- \dot \phi\sin(\theta)\cos(\phi)\cos(\theta) - \dot \theta \cos^2(\theta)\sin(\phi)\big)$$
$$= -mr^2\big(\dot \theta \sin(\phi) + \dot \phi \sin(\theta)\cos(\theta)\cos(\phi)\big)$$
$$L_y= m(z\dot x - x\dot z)= mr^2(\dot \theta \cos^2(\theta) \cos(\phi) - \dot \phi \sin(\theta)\cos(\theta)\sin(\phi) + \dot \theta\sin^2(\theta)\cos(\phi)\big)$$
$$= mr^2\big(\dot \theta \cos(\phi) -\dot \phi \sin(\theta)\cos(\theta) \sin(\phi)\big)$$
$$\implies -\sin(\phi)L_x + \cos(\phi)L_y = mr^2\dot\theta = p_\theta$$
as promised.
The Poisson bracket of the two angular momentum coordinates is then
$$\{p_\theta,p_\phi\} = \{\cos(\phi)L_y,L_z\} - \{\sin(\phi) L_x,L_z\}$$
Noting that $\{AB,C\}=A\{B,C\}+\{A,C\}B$ and that $\{f(\phi),L_z\} = \{f(\phi),p_\phi\} = f'(\phi)$, we quickly find that $\{p_\theta,p_\phi\}=0$ as expected.