Space forms are complete (connected) Riemannian manifolds of constant sectional curvature.
These fall into three classes: Euclidean, with universal covering isometric to $\mathbb{R}^n$,
spherical, with universal covering isometric to $S^n$, and hyperbolic, with universal covering isometric to $\mathbb{H}^n$.
Does there exist compact space forms of the same dimension from two different classes having isomorphic fundamental groups?
This cannot happen for $n = 2$, since the Gauss-Bonnet theorem
$$
\int_M K{ \ }\mathrm{vol}_M = 2{\pi}\chi(M)
$$
shows that the Euler characteristic $\chi(M)$ is positive, zero, or negative when the space form $M$ is Euclidean, spherical or hyperbolic respectively. But for (closed) surfaces the fundamental group determines the Euler characteristic.
It is essential for the question to require compactness, otherwise there are trivial examples. Dividing out $\mathbb{R}^2$ by the group of isometries generated by
$(x,y) \mapsto (x+1,y)$ yields a Euclidean space form with fundamental group isomorphic to $\mathbb{Z}$, while dividing out $\mathbb{H}^2$ by the group of isometries generated by
the hyperbolic isometry $(x,y) \mapsto (x+1,y)$ yields a hyperbolic space form also with fundamental group isomorphic to $\mathbb{Z}$.
The standard reference on space forms is Spaces of Constant Curvature by Joseph A. Wolf.
The classification of Euclidean space forms is given in Chapter 3, and of spherical ones in Chapter 7. Wolf does not treat hyperbolic space forms, possibly because not much was known about them in 1967. Unfortunately, the fundamental groups are infinite for the compact Euclidean space forms, and finite for the spherical space forms (which are necessarily compact, being quotients of $S^n$). So a hypothetical example has to involve a hyperbolic space form.
An example might drop out of the theory of three-manifolds. In dimension three the space forms belong to three of the eight Thurston model geometries. A pair yielding an example would have to be one Euclidean and one hyperbolic, since it follows from Perelman's geometrization theorem that the spherical ones are precisely those with finite fundamental group.
Best Answer
The fundamental group of a compact hyperbolic space form has exponential growth, according to a well-known theorem of Milnor [Milnor, J. A note on curvature and fundamental group. J. Differential Geometry 2 1968 1--7. MR0232311]. Bieberbach groups are, on the other hand, polynomial: indeed, their translations subgroups have finite index, so by Gromov's theorem they have polynomial growth.