I know since $[\hat{L}^2,\hat{L}_x] = 0$ and $[\hat{L}^2,\hat{L}_y] = 0$, they share the same eigenvectors. So this means spherical harmonics must be eigenvectors of $\hat{L}_x$ and $\hat{L}_y$.
What would $ \hat{L}_x\left| l,m\right>$ and $ \hat{L}_y\left| l,m\right>$ be? How would you go about finding the eigenvalues?
Shouldn't their eigenvalues also be the same because ladder operators $\hat{L}_+$ and $\hat{L}_-$ are conjugates of one another and are composed of $\hat{L}_x$ and $\hat{L}_y$?
I know since $[\hat{L}^2,\hat{L}_x] = 0$ and $[\hat{L}^2,\hat{L}_y] = 0$, they share the same eigenvectors.
This isn't right. The correct statement is that $[\hat L^2,\hat L_x]=0$ (along with the fact that $\hat L^2$ and $\hat L_x$ are both self-adjoint) implies that there exists an orthonormal basis $\mathcal B$ consisting of simultaneous eigenvectors of $L^2$ and $L_x$. It emphatically does not mean that every eigenvector of $\hat L^2$ is an eigenvector of $\hat L_x$ and vice-versa.
Similarly, $[\hat L^2,\hat L_y]=0$ implies that there exists an orthonormal basis $\mathcal B'$ consisting of simultaneous eigenvectors of $L^2$ and $\hat L_y$. There is no reason to expect that we could choose $\mathcal B=\mathcal B'$, and because $[\hat L_x,\hat L_y]\neq 0$ we are in fact guaranteed that such a choice is not possible.
So this means spherical harmonics must be eigenvectors of $\hat{L}_x$ and $\hat{L}_y$
No, for the reason stated above. The full set$^\ddagger$ of spherical harmonics (as conventionally defined) are eigenvectors of $\hat L^2$ and $\hat L_z$, not $\hat L_x$ or $\hat L_y$. Of course we could define rotated spherical harmonics which are eignevectors of e.g. $\hat L^2$ and $\hat L_x$, but they would not generally be eigenvectors of $\hat L_y$ or $\hat L_z$.
Shouldn't their eigenvalues also be the same because ladder operators $\hat{L}_+$ and $\hat{L}_-$ are conjugates of one another and are composed of $\hat{L}_x$ and $\hat{L}_y$?
I'm not really sure what you mean by this, but in the context of the above clarifications the question seems to become moot.
$^\ddagger$Individual spherical harmonics may be eigenvectors of $\hat L_x$ or $\hat L_y$ - specifically, $|0,0\rangle$ is an eigenvector of both with eigenvalue $0$ - but this does not constitute a full basis for the space. More generally, $[\hat A,\hat B]\neq 0$ does not generally mean that there are no simultaneous eigenvectors of $\hat A$ and $\hat B$ - only that there is no basis of such vectors.
By spherical symmetry (if your problem has spherical symmetry), the eigenvalues of $\hat L_x$ and $\hat L_y$ must (separately) be the same as those of $\hat L_z$.
However, the eigenvectors of $\hat L_z$, which are simultaneously eigenvectors of $L^2$, will not be eigenvectors of $\hat L_x$ or $\hat L_y$ since $[\hat L_z,\hat L_x]\ne 0$ and $[\hat L_z,\hat L_y]\ne 0$.
This is possible because, inside the set of those vectors with the same eigenvalue of $L^2$, one can construct separate sets which are also eigenvectors of one of $\hat L_z$, $\hat L_y$, $\hat L_z$.
It is of course possible to write eigenvectors of $\hat L_x$ as linear combination of eigenvectors of $\hat L_z$, because the eigenvectors of $\hat L_z$ form a complete basis inside the space of eigenvectors of $L^2$ with a given eigenvalue.
As remarked in another answer, the unique exception to this is the $L=0,M=0$ state, which is eigenvector of all the $\hat L_k$ with eigenvalue $0$ for all $\hat L_k$.
One way to verify that the eigenvalues of - say - $\hat L_y$ are the same as those of $\hat L_z$ is to write the matrix representation $\Gamma(\hat L_y)$ of $\hat L_y$ in the $2j+1$-dimensional space of states with a given $j$ value. From the matrix representation one finds the eigenvalues in the usual way: by solving for $\lambda$ in $\hbox{Det}(\Gamma(\hat L_y)-\lambda \hat 1)=0$.
Once you have the eigenvalues (either from the determinant or by just realizing they are the same as those of $\hat L_z$), you can search for the linear combinations of eigenvectors of $\hat L_z$ that will return each eigenvalue, i.e. by solving for the unknown coefficients $\alpha^k_m$ in \begin{align} \hat L_y\left(\sum_{m} \alpha^k_m Y_{\ell}^m\right)=\lambda_k \left(\sum_m \alpha^k_m Y_\ell^m\right)\, , \tag{1} \end{align} for the eigenvalue $\lambda_k$.
Eq.(1) will give you relations between the various $\alpha^k_m$’s, to eventually solve for each of them.
