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Thanks to Dave Benson for pointing out an algebra error in the first draft of this answer and a missing detail.

Suppose $X$ is homotopy equivalent to a finite dimensional, simply-connected, and noncontractible CW complex. Let $k=\mathbb{Q},\mathbb{Z}/p$. Then for one of these choices of $k$ there exists an $i>0$ such that $H_i(M; k)$ is nontrivial, otherwise $X$ would be contractible by the homology Whitehead theorem (see Corollary 3A.7.of Hatcher).

The Kunneth theorem implies $H_{2i}(M \times M;k)$ is nontrivial which implies $M$ can't be homotopy equivalent to $M \times M$.

The simply-connectedness assumption is necessary since the homology could potentially vanish in all degrees.

Thanks to Dave Benson for pointing out an algebra error in the first draft of this answer and a missing detail.

Suppose $X$ is homotopy equivalent to a finite dimensional, simply-connected, and noncontractible CW complex. Then for $k=\mathbb Q$ or $k = \mathbb Z/p$ for some $p$, there exists an $i>0$ such that $H_i(M; k)$ is nontrivial, otherwise $X$ would be contractible by the homology Whitehead theorem (see Corollary 3A.7 of Hatcher).

The Künneth theorem implies $H_{2i}(M \times M;k)$ is nontrivial which implies $M$ can't be homotopy equivalent to $M \times M$.

The simply-connectedness assumption is necessary since the homology could potentially vanish in all degrees.

Thanks to Dave Benson for pointing out an algebra error in the first draft of this answer.

Suppose $X$ is homotopy equivalent to a finite dimensional, simply-connected, and noncontractible CW complex. Let $k=\mathbb{Q},\mathbb{Z}/p$. Then for one of these choices of $k$ there exists an $i>0$ such that $H_i(M; k)$ is nontrivial, otherwise $X$ would be contractible by the homology Whitehead theorem.

The Kunneth theorem implies $H_{2i}(M \times M;k)$ is nontrivial which implies $M$ can't be homotopy equivalent to $M \times M$.

The simply-connectedness assumption is necessary since the homology could potentially vanish in all degrees.

Thanks to Dave Benson for pointing out an algebra error in the first draft of this answer and a missing detail.

Suppose $X$ is homotopy equivalent to a finite dimensional, simply-connected, and noncontractible CW complex. Let $k=\mathbb{Q},\mathbb{Z}/p$. Then for one of these choices of $k$ there exists an $i>0$ such that $H_i(M; k)$ is nontrivial, otherwise $X$ would be contractible by the homology Whitehead theorem (see Corollary 3A.7.of Hatcher).

The Kunneth theorem implies $H_{2i}(M \times M;k)$ is nontrivial which implies $M$ can't be homotopy equivalent to $M \times M$.

The simply-connectedness assumption is necessary since the homology could potentially vanish in all degrees.

Suppose $X$ is homotopy equivalent to a finite dimensional, simply-connected, and noncontractible CW complex. Then there exists a $k>0$ such that $H_k(M)$ is nontrivial. The Kunneth theorem implies $H_{2k}(M \times M)$ is nontrivial which implies $M$ can't be homotopy equivalent to $M \times M$.

The simply-connectedness assumption is necessary since the homology could potentially vanish in all degrees.

Thanks to Dave Benson for pointing out an algebra error in the first draft of this answer.

Suppose $X$ is homotopy equivalent to a finite dimensional, simply-connected, and noncontractible CW complex. Let $k=\mathbb{Q},\mathbb{Z}/p$. Then for one of these choices of $k$ there exists an $i>0$ such that $H_i(M; k)$ is nontrivial, otherwise $X$ would be contractible by the homology Whitehead theorem. 

The Kunneth theorem implies $H_{2i}(M \times M;k)$ is nontrivial which implies $M$ can't be homotopy equivalent to $M \times M$.

The simply-connectedness assumption is necessary since the homology could potentially vanish in all degrees.