 Solution

Take $x$ positive downward and put zero gravitational potential energy at the spring's support. The mass is then a distance $l+x$ below that reference, so $V_g=-mg(l+x)$. Hooke's law gives the elastic potential energy $k{x}^2/2$. ThereforeThe constant term $mgl$ fixes the requested reference but does not affect the motion. The Euler-Lagrange equation isThis is exactly Newton's second law with gravity downward and spring force $kx$ upward for positive extension. The static extension is $x_0=mg/k$.