I'll be working in the stable world. It's an easy observation that any 2-cell complex (over the sphere) with bottom cell in dimension zero is a Thom spectrum: any such complex is the cofiber of some element $\alpha\in \pi_n S$, so it is the Thom spectrum of the map $S^{n+1} \to B\mathrm{GL}_1 S$ classifying $\alpha$. Is there an example of a finite complex (again, with bottom cell in dimension zero, which is a necessary condition) which is *not* a Thom spectrum? Is there such a 3-cell complex?

Edit: I think the question mark complex $Q$ gives an example of such a 3-cell complex. This is the complex constructed by lifting the stable map $\Sigma \eta:S^2\to S^1$ to $\Sigma^{−1}\mathbf{R}P^2$ by choosing a nullhomotopy of $2\eta$; it follows that if $Q$ was to be a Thom spectrum, it would be a Thom spectrum of bundle over a space with bottom cell in dimension 1 and top cell in dimension 3, along with a $\mathrm{Sq}^2$ in cohomology. But $\mathrm{Sq}^2$ (unstably) vanishes on any cohomology class in dimension 1, so this is impossible.

I guess, then, my question could be revised to asking whether there are nontrivial sufficient conditions under which a finite complex can be realized as a Thom spectrum.

rank 0stable spherical fibration (since they mention of $BGL_1(\mathbb{S})$ and not $BGL_1(\mathbb{S})\times \mathbb{Z}$), so you're not allowed to desuspend. $\endgroup$ – Denis Nardin Jan 13 '19 at 20:58