Consider the set $\displaystyle S=\{p(x) \in P_m[x]\: : \: p(2)=0\}$. Clearly this is a subspace of $\displaystyle P_m[x]$, and every $\displaystyle p(x) \in S$ can be written as $\displaystyle (x-2)g(x)$. Consider the map

$\displaystyle T : S \rightarrow P_{m-1}[x]$

$\displaystyle (x-2)g(x) \mapsto g(x)$

This is a surjective (i.e. onto) linear transformation, because to every $\displaystyle g(x) \in P_{m-1}[x]$, we have $\displaystyle T((x-2)g(x))=g(x)$. Moreover it is injective (i.e. one-to-one) because the inverse image of the zero polynomial is just the zero polynomial (check this). Hence $\displaystyle \mbox{dim }S = m$.

It follows that since you have $\displaystyle m+1$ elements of $\displaystyle S$ in your list, they cannot be linearly independent.