The definition of the adiabatic index of a relativistic polytrope fluid is this: $$\tag{1} \gamma = \frac{\rho + p}{p} \, \frac{dp}{d\rho}, $$ where $\rho$ is the energy density (not the mass density). The sound velocity in the fluid is restricted by the principle of causality, so (I'm using units such that $c \equiv 1$): $$\tag{2} v_s^2 \equiv \frac{dp}{d\rho} \le 1 \qquad \Rightarrow \qquad \gamma \le \frac{\rho + p}{p}. $$ For a constant $\gamma$, one solution to (1) is non-linear: $$\tag{3} \rho = \frac{p}{\gamma - 1} + (p / \kappa)^{1/\gamma}, $$ where $\kappa$ is a positive integration constant. Then, a linear EoS (equation of state) is another solution for the same value of $\gamma$: $$\tag{4} \rho = \frac{1}{\gamma - 1} \, p. $$ Clearly, this linear function has $v_s > 1$ for $\gamma > 2$.
In the special case of $\gamma = 2$, (3) can be inverted: $$\tag{5} p(\rho) = \frac{1}{4 \kappa} \bigl( \sqrt{1 + 4 \kappa \rho} - 1 \Bigr)^2. $$ This non-linear EoS gives $p(\rho) \approx \kappa \rho^2$ when $4 \kappa \rho \ll 1$, and gives $p(\rho) \approx \rho$ when $4 \kappa \rho \gg 1$. The speed of sound is always lower than 1 (except in the limiting case $\rho \rightarrow \infty$): $$\tag{6} v_s^2 = \frac{dp}{d\rho} = 1 - \frac{1}{\sqrt{1 + 4 \kappa \rho}}. $$ This could also be the case for $\gamma > 2$, when the EoS is non-linear, so from (1)-(3), how can we prove that $v_s \le 1 \; \Rightarrow \; \gamma \le 2$ even for a non-linear EoS?