= Solution
\b[The assertion is false.] Take $m=3$ and $n=2$. For the constant attaching map, the <cell attachment> gives $S^2\vee S^4$. Its degree-two generator has square zero: restricting the square to either sphere gives zero, and restrictions identify its degree-four <cohomology> with that of the $S^4$ summand.
For the <Hopf fibration> $\eta:S^3\to S^2=\mathbb{CP}^1$, the attachment instead gives $\mathbb{CP}^2$. One can verify the attaching map explicitly with the characteristic map
$$
D^4\longrightarrow\mathbb{CP}^2,\qquad (z_1,z_2)\longmapsto[z_1:z_2:\sqrt{1-|z_1|^2-|z_2|^2}].
$$
Its interior maps homeomorphically to the complement of $\mathbb{CP}^1$; on the boundary it sends $(z_1,z_2)$ to $[z_1:z_2:0]$, exactly the <Hopf fibration>. By part (a), the degree-two generator of $H^*(\mathbb{CP}^2;\mathbb Z)$ has nonzero square generating degree four. Hence
$$
\boxed{H^*(X_{\mathrm{constant}};\mathbb Z)\not\cong H^*(X_\eta;\mathbb Z)\text{ as graded rings}.}
$$
The additive groups agree, but multiplication distinguishes the attachments.
The dimension condition explains why this example is the relevant one. In general the only positive-degree additive generators lie in degrees $n$ and $m+1$. A potentially nonzero product can only be the square of the degree-$n$ generator, and only when $m+1=2n$. Its coefficient is the <Hopf invariant> of the attaching map. Here the constant map has invariant zero, whereas the complex <Hopf fibration> has invariant one.
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