if we look at the Lagrangian for the Kepler problem in cartesian coordinates, we obtain sth like
$$\mathcal L\left( \vec X, \vec x, \dot{\vec{X}}, \dot{\vec{x}} \right) = \frac{M}{2} \dot{\vec{X}}^2 + \frac{m}{2} \dot{\vec{x}}^2 + \frac{GMm}{\left| \vec x \right|},$$
where $\vec x$ describes the relative motion and $\vec X$ the motion of the center of mass.
$$ M=m_1+m_2, \qquad m = \frac{m_1 m_2}{m_1+m_2}. $$ Now, applying the Lagrange equations of the second kind, we obtain
$$M\ddot{\vec{X}} = 0 \\ m\ddot{\vec{x}} = \vec\nabla \left( \frac{GMm}{\left| \vec x \right|} \right) = -\frac{GMm}{\left|\vec x\right|^3}\vec{x}$$
Now, solving the equation for the center of mass is very easy, but can we actually solve the second differential equation in cartesian coordinates? Because if not, then I understand why we move to spherical coordinates for the Lagrangian, at least this is what my textbook does now.