Dear asmaier, sometimes it's useful to look into the real world to avoid some simple mistakes. The actual speed of sound in the air is 340 meters per second which is about one millionth of the speed of light. The squared speed of sound is one trillionth of the squared speed of light, so your claim that the non-relativistic speed of sound is comparable to the speed of light is surely incorrect, isn't it?
The first formula by Weinberg that you quoted is universally valid but you apply it incorrectly. Well, Weinberg doesn't make these errors so that his first $\sqrt{1/3}$ result for the relativistic gas is correct, too. However, in particular, $(\rho-nm)$ is supposed to measure the energy above the latent energy of $E=mc^2$, because only this "purely kinetic energy" contributes to the pressure; it must still be multiplied by $(\gamma-1)$.
Clearly, $(\rho-nm)$ is negligibly small in the non-relativistic limit. So your $c\sqrt{2/3}$ for the nonrelativistic speed of sound is clearly invalid. In fact, $(\rho-nm)$ measures the kinetic energy of the molecules, $mv^2/2$; the factor of $m$ cancels. Because it will get differentiated and multiplied by $(\gamma-1)$, and we take the square root at the end, you may see that the speed of sound comes out approximately equal to the speed of molecules in the air, just slightly smaller than that (mainly because of the factor of $1/2$ in the kinetic energy $mv^2/2$).
Recall that the (root mean square) speed of air molecules is 500 meters per second and the speed of sound is 340 meters per second.
There is no contradiction between the relativistic formulae and the non-relativistic ones, which are the limits of the former, and the non-relativistic materials' speed of sound is vastly smaller than the speed of light.
To find the possible waves in a solid you really need the full theory tensor theory of elasticity. The transverse shear wave is the simplest case though. This is because, unlike the general elastic response, the force is parallel to the transverse displacements. You can find the wave equation by just equating the force gradient
$$
\frac {\partial F_y} {\partial x}=\mu \frac {\partial^2 y}{\partial x^2}
$$
to the acceleration $y$ component
$$
\rho \frac {\partial^2 y}{\partial t^2}.
$$
Thus $c^2_{\rm transverse}= \mu/\rho$.
Here $\mu$ is the shear modulus. (This analysis does require that you have an infinite plane wave)
Best Answer
Yes, you're wrong.
Sound waves are small compressions (oscillations) of an elastic medium, travelling through that same elastic medium (as a wave). Air, liquids or solids are typical elastic media through which sound waves can travel.
Vacuum however contains no matter and cannot sustain sound waves at all.
Watch this video on a bell in a vacuum jar.