It suffices prove that if $P(x)$ is a polynomial with exactly $k\in\mathbb N$ non-zero coefficients, then $x=1$ is not a root of $P(x)$ with multiplicity $\ge k$. This can be done by induction on $k$. For $k=1$, this is trivial. Suppose that the hypothesis statement holds for some $k\in\mathbb N$. Fix some $P(x)$ with exactly $k+1$ non-zero coefficients. WLOG, the constant term of $P(x)$ is nonzero (because we can divide $P(x)$ by proper $x^n$.) Now, suppose to the contrary that $x=1$ is a root of $P(x)$ with multiplicity $\ge k+1$. Then $x=1$ is a root of $P'(x)$ with multiplicity $\ge k$. However, $P'(x)$ has exactly $k$ non-zero terms, which contradicts to the hypothesis statement for $k$. Therefore, the hypothesis statement also holds for $k+1$. So the hypothesis statement holds for all positive integer $k$.

Another approach. A pinch of linear algebra is used.