= Solution
\b[True for two positive-dimensional factors.] In fact degree three can be replaced by any nonzero <mapping degree>. Suppose $M=X^k\times Y^{m-k}$ with $0<k<m$. The factors are closed and connected. They are also orientable: an orientation-reversing loop in either factor would reverse the product <orientation> while the other factor stays fixed, contradicting the orientation of $M$.
Let $\omega_X,\omega_Y$ be their integral top <orientation classes>, and let $\alpha,\beta$ be the corresponding projection pullbacks to $M$. The <Künneth theorem> and the product <fundamental class> give
$$
\alpha\smile\beta=\pm\omega_M\in H^m(M;\mathbb Z).
$$
For $f:S^m\to M$, both $f^*\alpha$ and $f^*\beta$ vanish, since a sphere has no <cohomology> in those intermediate positive degrees. Naturality of the <cup product> therefore gives
$$
f^*\omega_M=\pm f^*\alpha\smile f^*\beta=0.
$$
On the other hand, the definition of degree three gives $f^*\omega_M=3\omega_{S^m}\ne0$. This contradiction proves the <nonzero degree from a sphere obstructs manifold products> assertion. \b[The product with a zero-dimensional point factor is the excluded trivial case.]
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