Example of two spaces indisinguishable by their homology modules (with $\mathbb{Z}$ coefficients) but with different cohomology rings I'm running a student seminar on cohomology (for masters students) and would like to motivate the dualisation of homology by talking about cup products. So I'm looking for an example of two spaces $X$ and $Y$ with the same homology modules but different cohomology rings. Are there any nice-ish examples which would be reasonable to talk about in a seminar?
I do already have the example of $X=\mathbb{R}P^n$ and $Y=\vee_{i\leq n}S^i$ with $\mathbb{Z}/2$ coefficients. The problem with this is that these spaces are distinguished by their homology with $\mathbb{Z}$ coefficients. This might still be sufficient motivation for this kind of seminar, but I'd still prefer to have an example of two spaces where you really need the extra ring structure to tell them apart.
Thanks for any help

This simple example may be outside the scope of your seminar, but it adds a layer of complexity to the other examples already given.
Consider $\Sigma \mathbb{C}P^n$ and $\bigvee_{i=1,\dots n} S^{2i+1}$. These spaces have the same cohomology groups for all coefficients, and indeed the same cohomology rings, since all cup products in the suspension  $\Sigma \mathbb{C}P^n$ are trivial. 
How can cohomolgoy distinguish between the homotopy types of these spaces? Their mod $p$ cohomology rings are not isomorphic as modules over the Steenrod algebra, since $\Sigma \mathbb{C}P^n$ will admit non-trivial operations for suitable $p$ whilst $\bigvee_{i=1,\dots n} S^{2i+1}$ will not. In particular, if $x\in H^2(\mathbb{C}P^2\mathbb{Z}_2)$ is a generator then so is $Sq^2x=x^2\in H^4(\mathbb{C}P^2;\mathbb{Z}_2)$. Thus if $\sigma$ denotes the supension isomorphism, which commutes with the Steenrod operations, then $Sq^2(\sigma x)=\sigma(Sq^2x)=\sigma (x^2)\in H^5(\Sigma\mathbb{C}P^2;\mathbb{Z}_2)$ is non-trivial.
