From the Künneth theorem you can check that there is no class $\omega\in H^2(M^{2d}\times S^{2n};\mathbb{R})$ such that $\omega^{d+n}\neq 0$. (This is an excellent thing for you to work out for yourself.) Since a symplectic form is closed and nondegenerate, this shows that no symplectic structure on $M\times S^{2n}$ can exist for $n>1$ if $M$ is compact.

(Thanks to Eric and David for pointing out that I was assuming $M$ to be compact; this is necessary so that nondegeneracy implies that the top power of $\omega$ represents a nonzero multiple of the fundamental class. In his answer David Speyer gives a nice counterexample if $M$ is not required to be compact.)

From the Künneth theorem you can check that there is no class $\omega\in H^2(M^{2d}\times S^{2n};\mathbb{R})$ such that $\omega^{d+n}\neq 0$. (This is an excellent thing for you to work out for yourself.) Since a symplectic form is closed and nondegenerate, this shows that no symplectic structure on $M\times S^{2n}$ can exist for $n>1$ if $M$ is compact.

(Thanks to Eric and David for pointing out that I was assuming $M$ to be compact; this is necessary so that nondegeneracy implies that the top power of $\omega$ represents a nonzero multiple of the fundamental class. In his answer David Speyer gives a nice counterexample if $M$ is not required to be compact.)

From the Künneth theorem you can check that there is no class $\omega\in H^2(M^{2d}\times S^{2n};\mathbb{R})$ such that $\omega^{d+n}\neq 0$. (This would be an excellent thing for you to work out for yourself.) Since a symplectic form is closed and nondegenerate, this shows that no symplectic structure on $M\times S^{2n}$ can exist for $n>1$.

From the Künneth theorem you can check that there is no class $\omega\in H^2(M^{2d}\times S^{2n};\mathbb{R})$ such that $\omega^{d+n}\neq 0$. (This is an excellent thing for you to work out for yourself.) Since a symplectic form is closed and nondegenerate, this shows that no symplectic structure on $M\times S^{2n}$ can exist for $n>1$ if $M$ is compact.

(Thanks to Eric and David for pointing out that I was assuming $M$ to be compact; this is necessary so that nondegeneracy implies that the top power of $\omega$ represents a nonzero multiple of the fundamental class. In his answer David Speyer gives a nice counterexample if $M$ is not required to be compact.)