There are no such functions.

Suppose $f\in L^2(\mathbb R)$ is supported in a set $S$ of finite measure and assume its Fourier transform $g=\hat f$ satisfies $|g(x)|\leq Ae^{-k|x|^\gamma}$ for $k>0$ and $\gamma>1$.
Denote $I_m=[-m,m]$.

By [Benedick's theorem](http://en.wikipedia.org/wiki/Uncertainty_principle#Benedicks.27s_theorem) there is a constant $C>0$ so that
$$
\|f\|_{L^2(\mathbb R)}
\leq
Ce^{2Cm|S|}\|g\|_{L^2(\mathbb R\setminus I_m)}.
$$
If $m\geq\gamma^{-1/(\gamma-1)}$, then $x\geq m$ implies $x^\gamma-x\geq m^\gamma-m$.
Therefore
$$
\|g\|_{L^2(\mathbb R\setminus I_m)}^2
=
\sum_\pm\int_m^\infty|g(\pm x)|^2dx
\leq
\sum_\pm\int_m^\infty A^2e^{-2kx^\gamma} dx
\leq
2A^2e^{-2km^\gamma}\int_m^\infty e^{-2k(x-m)} dx
=
2A^2e^{-2km^\gamma}(2k)^{-1}.
$$
Inserting this to the previous estimate gives
$$
\|f\|_{L^2(\mathbb R)}
\leq
Ck^{-1/2}Ae^{2Cm|S|-km^\gamma}.
$$
Since $\gamma>1$, taking the limit $m\to\infty$ gives $\|f\|_{L^2(\mathbb R)}=0$ so $f=g=0$.