I asked this question on MSE 9 days ago and it got a very helpful comment from Eric Towers providing the Palais Stewart reference, but no answers. So I'm crossposting it here.

Let 
$$
T^n \to M \to T^m 
$$
be a principal torus bundle over a torus. Then $ M $ is a solvmanifold, even a nilmanifold (in fact $ M $ is the total space of a principal torus bundle over a torus if and only if it is a compact nilmanifold for a 2 step nilpotent Lie group, this is theorem 3 of [Palais, Stewart, TORUS BUNDLES OVER A TORUS]). 

What if the bundle is not necessarily principal? Is every torus bundle over a torus a solvmanifold? In other words, if we have a fiber bundle
$$
 T^n \to M \to T^m 
$$
then can we conclude that the total space $ M $ is a solvmanifold? 

EDIT: 

The answer, comments and references from Igor Beledrek prove that something much stronger is true: A manifold $ M $ is homeomorphic to a compact solvmanifold if and only if $ M $ is the total space of a bundle 
$$
N \to M \to T^n
$$
where $ N $ is a compact nilmanifold and $ T^n $ is a torus.

The smooth case is also addressed. The answer, comments and references from Igor Beledrek demonstrate that already in dimension 4 there exists a smooth manifold $ M $ (in particular an exotic 4 torus)
$$
T^2 \to M \to T^2
$$
which arise as the total space of a torus bundle over a torus (both base and fiber with standard smooth structure) but which is not diffeomorphic to any solvmanifold. So in every dimension $ d \geq 4 $ there are smooth torus bundles over a torus that are not diffeomorphic to any solvmanifold.

For $ d \leq 3 $ we have by Moise's theorem that there are no exotic smooth structures. So all torus bundles over a torus with total space of dimension $ d \leq 3 $ are diffeomorphic to a solvmanifold.