Actually, there are two exterior Schwarzschild solutions (the name vacuum solution is traditionally reserved for $T_{\mu \nu}$ identically zero). If one assumes $p=0$, then from Einstein equations it follows that $\varepsilon=0$, too. On the other hand, if one assumes $\varepsilon=0$, it does not imply necessarily that $p=0$. The second solution is unlike the first, not asymptotically flat. Thus, it is a cosmological solution. It describes spacetime without energy density, but with pressure. The question whether such exterior metric would explain galaxy rotational curves would need a closer investigation. I speculate here: a spacetime without matter but with a non-vanishing mean hydrostatic stress (=pressure) could be an interesting new candidate for the mysterious dark matter.
Derivation
The line element in static spherically-symmetric space-time can be written as
\begin{equation}
\label{metric}
{\rm d}s^2={\rm e}^{2\nu}c^2{\rm d}t^2-{\rm e}^{2\lambda}{\rm d}r^2-r^2{\rm d}\Omega^2~,
\end{equation}
with curvature radius $r$ and the infinitesimal surface element ${\rm d}\Omega$ of a 2-sphere.
The metric components satisfy Einstein field equations (EFE)
\begin{equation}\label{einstein}
{\rm{R}}_{\mu}^{~\nu}-\frac{1}{2}{\rm{R}}_{\lambda}^{~\lambda}~\delta_{\mu}^{\nu}=\kappa {\rm{T}}_{\mu}^{~\nu},
\end{equation}
where, in case of a perfect fluid, the stress energy tensor has diagonal form ${\rm{T}}_{\mu}^{\nu}\equiv {\rm{diag}}~\{\varepsilon,-p,-p,-p\}$. Multiplying both sides of EFE equation with square of curvature radius of fluid sphere $R^{2}$ makes this equation dimensionless. The dimensionless pressure and energy density in subsequent considerations are $\varepsilon\equiv {\varepsilon}~{\kappa}~c^2R^2$ and $p\equiv {p}~{\kappa}~R^2$, with $\kappa$, the Einstein's gravitational constant $8\pi G/c^4$. The metric components functions $\nu$ and $\lambda$ depend only on the dimensionless variable $u\equiv r^2/ R^2$ and compactness parameter $\alpha\equiv r_S/R$.
The requirement for isotropic pressure reduces Einstein's field equations to these three differential equations
\begin{equation}\label{compactform}
{\rm e}^{-\lambda}\frac{\rm d}{{\rm d} u} \left( {\rm e}^{-\lambda}\frac{\rm d}{{\rm d}u}{\rm e}^{\nu}\right) = \frac{1}{4}\frac{\rm d}{{\rm d}u}\left(\frac{1-{\rm e}^{-2\lambda}}{u}\right)~{\rm e}^{\nu}~,\tag{1}
\end{equation}
\begin{equation}\label{pressure}
p=\frac{4}{{\rm e}^{\nu}}\frac{{\rm d}{\rm e}^{\nu}}{{\rm d}u}{\rm e}^{-2\lambda}-\frac{1-{\rm e}^{-2\lambda}}{u}~,\tag{2}
\end{equation}
\begin{equation}
\label{density}
\varepsilon=\frac{1-{\rm e}^{-2\lambda}}{u}-2~\frac{{\rm d}{\rm e}^{-2\lambda}}{{\rm d}u}~.\tag{3}
\end{equation}
The first equation can be read as a first-order linear non-homogeneous equation for ${\rm e}^{-2\lambda}$, or as a second-order homogeneous linear differential equation for ${\rm e}^{\nu}$. Therefore, in total there are three boundary conditions which can be set on the star surface ($u=1$). They express continuity of metric and requirement for total mass $M$ being enclosed by spherical surface of area $4\pi R^{2}$. The boundary condition for ${\rm e}^{-2\lambda}$ is
\begin{equation}\label{BC1}
{\rm e}^{-2\lambda}(1,\alpha)=1-\alpha~,\tag{4}
\end{equation}
and the two boundary conditions for ${\rm e}^{\nu}$ are
\begin{equation}\label{BC2&3}
{\rm e}^{\nu}(1,\alpha)=\sqrt{1-\alpha},\tag{5}
\end{equation}
\begin{equation}
\frac{{\rm d}{\rm e}^{\nu}}{{\rm d}u}(1,\alpha) = \frac{1}{4}\frac{p_{1}+\alpha}{\sqrt{1-\alpha}}~,~~~p_{1}\equiv{p(1,\alpha)}.\tag{6}
\end{equation}
The boundary condition (6) in not independent. It results from equation (2) by stetting (4) and (5) into it.
Let's solve EFE for static spacetime with a sphere of constant energy density
\begin{equation}
\label{Schwarzschildsolution}
\varepsilon =\left\{ \begin{array} {rcl} 3~\alpha & \mbox{for} & 0 \leq u \leq 1 \\\\ 0 & \mbox{for} & 1 > u \leq \infty~. \end{array}\right.\tag{7}
%\right\parskip
\end{equation}
The solution of differential equation (3) for metric function $\rm{e}^{-2\lambda}$ reads
\begin{equation}
\label{e-2lambda}
{\rm e}^{-2\lambda} =\left\{ \begin{array}{rcl} 1-\alpha~u~ & \mbox{for} & 0\leq u \leq 1
\\
\\1-\alpha/\sqrt{u}~ & \mbox{for} & 1 > u \leq \infty~. \end{array}\right.\tag{8}
\end{equation}
The solution of equation (1) yields
\begin{equation}
\label{enulambdazero}
{\rm e}^{\nu} =\left\{\begin{array}{rcl} (3/2+p_{1}/\alpha)~\sqrt{1-\alpha }-(1/2+p_{1}/\alpha)~\sqrt{1-\alpha u } & \mbox{for} & 0\leq u \leq 1
\\
\\
\Bigl(1-p_{1}/8~f(1,\alpha)\Bigr)~\sqrt{1-\alpha/\sqrt {u}}+p_{1}/8~\sqrt{1-\alpha}~f(u,\alpha)& \mbox{for} & 1 < u \leq \infty~. \end{array}\right.\tag{9}
\end{equation}
where auxiliary function $f$ is defined as
\begin{equation}
f(u,\alpha)\equiv{15~\alpha^{2}~\sqrt{1-\alpha/\sqrt{u}}~\tanh^{-1}{\sqrt{1-\alpha/\sqrt{u}}}-15~\alpha^{2}+5\alpha\sqrt{u}+2u}.\tag{10}
\end{equation}
Because for large $u$, $f(u)$ behaves like $\sim 2u$, an asymptotically flat solution is possible only if $p_{1}$ is zero. In that case the metric reduces to the well-known interior and exterior Schwarzschild metric solution.
The special case - Universe without matter but with pressure
Setting $\alpha$ to zero ($\varepsilon=0$) one obtains metric
\begin{equation}
\label{metricwithoutmatter}
{\rm d}s^2=(1-p_{1}/4+p_{1}/4~u)^2~c^2{\rm d}t^2-{\rm d}r^2-r^2{\rm d}\Omega^2~,\tag{11}
\end{equation}
where $p_{1}\equiv p(R)$ and $u\equiv (r/R)^{2}$.
The pressure and energy density are given as
\begin{equation}
\label{metricwithoutmatterpressure}
p=p_{1}~{\rm e}^{-\nu}\equiv p_{1}/(1-p_{1}/4+p_{1}/4~u),~~~~~~\varepsilon=0.\tag{12}
\end{equation}
For $p_{1}=4$, the metric (11) reduces to
\begin{equation}
\label{metricwithoutmatterpressurespecial}
{\rm d}s^2=(r/R)^4~c^2{\rm d}t^2-{\rm d}r^2-r^2{\rm d}\Omega^2~,\tag{13}
\end{equation}
and the pressure and energy density are
\begin{equation}
\label{metricwithoutmatterpressure2}
\varepsilon=0,~~~~~~p=2~c^{4}/G\cdot 1/(4\pi r^{2}).\tag{14}
\end{equation}
This solution could be possibly interpreted as a black hole without mass.
