Let $deg(P)=n>0$ and assume without loss of generality that $deg(Q) \leq deg(P)$. Consider the polynomial $R=(P-Q)P'$ ($P'$ denote the derivative of $P$). We have :
$$deg(R)\leq 2n-1 $$
Now if $r$ is a root of multiplicity $k$ of $P$ then $r$ is a root of $P'$ of multiplicity $k-1$ and because $Q(r)=0$, $r$ is a root of $P-Q$. hence $r$ is a root of $R$ of multiplicity at least $k$. So the n zeros of $P$ produces at least $n$ zeros of $R$, when multiplicities are counted.
The same pattern can be applied to $P-1$, every root $r$ of multiplicity $k$ of $P-1$ is a root of $P-Q$ and a root of multiplicity $k-1$ of $(P-1)'=P'$. So the n zeros of $P-1$ produces at least another $n$ zeros of $R$ when multiplicities are counted.
(the zeros of $P$ are all different with the zeros of $P-1$ it's obvious).
This means that $R$ has at least $2n$ roots and beacuse $deg(R)\leq 2n-1 $ we conclude that $R=0$ which implies $P=Q$.