= Solution
At zero an <ordinary point> requires $p,q$ to be analytic. A <regular singular point> requires $zp$ and $z^2q$ to extend analytically to zero, with at least one of $p,q$ failing to be analytic if “singular” excludes <ordinary points>.
Put $\zeta=1/z$ and $v(\zeta)=w(1/\zeta)$. Differentiation gives $w'=-\zeta^2v'$ and $w''=\zeta^4v''+2\zeta^3v'$, so the transformed equation is
$$
v''+\left(\frac2\zeta-\frac{p(1/\zeta)}{\zeta^2}\right)v'
+\frac{q(1/\zeta)}{\zeta^4}v=0.
$$
The point at infinity is ordinary precisely when both transformed coefficients are analytic at $\zeta=0$. Solving these conditions for the original coefficients gives
$$
\boxed{p(z)=2z^{-1}+z^{-2}P(z^{-1}),\qquad q(z)=z^{-4}Q(z^{-1})},
$$
with $P,Q$ analytic near zero; the sign can be absorbed into the arbitrary analytic $P$.
If the only singular point on the sphere is the regular singularity at zero, then $zp(z)$ and $z^2q(z)$ are entire. The condition at infinity makes the first tend to $2$ and the second to $0$. By <Liouville theorem> they are respectively the constants $2$ and $0$. Thus the unique equation is
$$
\boxed{w''+\frac2z w'=0},
$$
with general solution $w=A+B/z$. Zero is genuinely regular singular and infinity is ordinary. If “no other singular points” were restricted to finite points only, the general coefficients would instead be $p=P(z)/z$ and $q=Q(z)/z^2$ with entire $P,Q$; the preceding question about infinity indicates the spherical interpretation.
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