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Yes, the eigenvectors found with this method may depend on $x$ and $y$, but no, it doesn't matter in practice. If $A$ and $B$ share a basis of common eigenvectors, then $$ A = V\operatorname{diag}(\alpha_1,\dots,\alpha_n)V^{-1}, \quad B = V\operatorname{diag}(\beta_1,\dots,\beta_n)V^{-1}. $$ If $A$ has all distinct eigenvalues, then $V$ is unique, up to ...

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