 Solution

For $a,b>0$, dilation by $t=(b/a{)}^{1/m}$ maps $X_a$ to $X_b$, because $p(tx)=t^{m}p(x)=b$. Its inverse is dilation by $t^{-1}$, so all nonempty positive level manifolds are diffeomorphic.

A negative level need not be diffeomorphic to a positive one, even when both are nonempty. For $p(x,y,z)=x^2+y^2-z^2$, $X_1$ is connected (parametrize it by $(\sqrt{1+z^2}\cos \theta ,\sqrt{1+z^2}\sin \theta ,z)$), whereas $X_{-1}$ has two components, according to $z=\pm \sqrt{1+x^2+y^2}$. Thus the answer for opposite signs is no. If $m$ is odd, the additional symmetry $x\mapsto -x$ does interchange opposite levels up to scaling, but the question allows even degrees.