Can one hear the (topological) shape of a drum?

There are examples due to Ikeda of isospectral Lens spaces which are not homotopy equivalent.

Likeliest the simplest examples are the compact connected 3-dimensional flat manifolds which are a tetracosm and didicosm. These are isospectral but not homotopy equivalent. I'm not sure if they are known to have the same spectrum for the Laplace-de Rham operator (i.e. on forms). Presumably this could be computed explicitly.

There are many other examples of non-isometric manifolds which are strongly isospectral (having the same spectrum on forms) using Sunada's method, and arithmetic techniques (originally due to Vignéras) for locally symmetric spaces. Coupled with the Mostow rigidity theorem for locally symmetric spaces, this implies that these manifolds are not homotopy equivalent.

In 1-dimension, the spectrum determines the manifold.

For connected two-dimensional surfaces, the Laplace-de Rham spectrum determines the betti numbers, and hence the topological type (the surface $S$ is orientable iff $b_2(S)=1$, and then $b_1(S)$ determines the topological type if it is orientable or non-orientable). (For surfaces with boundary, an example of Bérard-Webb shows that the Neumann spectrum does not determine orientability.) Thus, the 3-dimensional examples coming from Sunada's construction or Vignéras' are minimal dimensional examples which are strongly isospectral. I don't know if the Laplace spectrum determines the topological type of closed surfaces.

See section 5 of the survey paper by Carolyn Gordon for more information on how topology cannot be detected by the spectrum and references. Note also that Lubotzky-Samuels-Vishne showed that there are isospectral arithmetic lattices which are not commensurable (so not obtained by Sunada's method or Vigneras' criterion).