This is a slightly expanded (and slightly more systematic) summary of my comments above. First of all, the equation
$$
-\varphi''+\frac{1}{r}\varphi' + (V+\frac{m}{r^2})\varphi=\lambda\varphi \quad\quad\quad\quad (1)
$$
fits into the standard Sturm-Liouville theory if written in the form
$$
T\varphi\equiv-\left( \frac{1}{r}\varphi'\right)'+W\varphi =\frac{\lambda}{r}\varphi ,
$$
with $W=V/r+m/r^3$. Now $T$ defines a symmetric operator on $L^2(I,dr/r)$.
The equation can be brought to Schrodinger form by the substitution $y=r^{-1/2}\varphi$. Then (1) becomes
$$
-y'' + \left( V + \frac{m+3/4}{r^2} \right) y = \lambda y .\quad\quad\quad\quad (2)
$$
Now the LHS defines an operator $Sy$ on $L^2(I, dr)$ which is unitarily equivalent to $T$.
Near $r=0$, this is a bounded perturbation of the $c/r^2$ potential. For $c\ge 3/4$, as here, we have limit point case: For $\lambda=0$ (and $V=0$), we can solve explicitly by $y=r^{\alpha}$, with $\alpha=1/2\pm \sqrt{1/4+c}$, and for $c\ge 3/4$, only the solution with $+$ is in $L^2$ near $r=0$.
As a consequence, the minimal operator (the closure of $S$ on $C_0^{\infty}(0,1)$) is symmetric with deficiency $(1,1)$. The self-adjoint realizations are obtained by imposing a boundary condition at $r=1$ only on the elements of the domain of the adjoint. The general boundary condition is $y(1)\cos\beta +y'(1)\sin\beta =0$; this gives all self-adjoint realizations.
As for the spectrum, I doubt that one can find this explicitly even when $V=0$. If $V$ is analytic, then $r=0$ is a regular singular point of (2), so the solution that is in $L^2$ will be of the form $y(r,\lambda)=r^{\alpha}\sum_{n\ge 0}c_n(\lambda)r^n$, with $\alpha=1/2+\sqrt{1+m}$ from above. The eigenvalues will be the solutions of $y(1,\lambda)=0$. (In particular, the regularity of these solutions proves that the spectrum is purely discrete, which need not hold in general when there is a limit point endpoint.)
