You are right. $|l=0,m=0\rangle$ (i.e., $Y^0_0$ ) is a simultaneous eigenvector of $L^2$ and $L_x,L_y,L_z$ with eigenvalue $0$. What is impossible is the existence of a common basis of eigenvectors of $L_x$ and $L_y$: the two operators would be diagonal with respect to that common basis and therefore they would commute in contrast to the commutation relation you wrote since $L_z\neq 0$.
With the same argument you see that $L_z\psi=0$ if $\psi \neq 0$ is a simultaneous eigenvector of $L_x$ and $L_y$. So $\psi$ is also an eigenvector of $L_z$ with eigenvalue $0$. Permuting the axes, the same argument proves that $L_x\psi=L_y\psi=0$. Notice that such eigenvector is also an eigenvector of $L^2$. As a consequence, there are no common eigenvalues when $L^2$ assumes semi-integer values. In fact, it is known that the eigenvalues pf $L^2$ (I am referring to it as a general angular momentum observable, not necessarily the orbital momentum) are of the form $\hbar l(l+1)$, where $l$ is integer $l= 0,1,2, \ldots$ or semi integer $l= 1/2, 3/2, 5/2,\ldots$. Correspondingly, the eigenvalues $m$ of $L_z$ take values $-l, -l+1, ..., l-1, l$. Therefore, $m$ may be $0$ only if $l$ is integer.
Summing up, assuming that $L_x,L_y,L_z$ represent the total angular momentum of a given physical system (including the spin if there is),
- There are no common (orthonormal) basis of eigenvectors of $L_x,L_y$;
- If a common eigenvector exists for $L_x,L_y$ it must have eigenvalue $0$ simultaneously for $L_x,L_y,L_z$ and also $L^2$;
- A common eigenvector of $L_x,L_y$ exists if and only if the operator $L^2$ -- whose eigenvalues are always in the set $\hbar l(l+1)$ for $l \in \mathbb N/2$ -- has eigenvalues of the form $l=k/2$ with even $k$.