Much more is true. According to [the answer][1] by Alexandre Eremenko (I do not have this paper, but I completely trust him), from $P(\mathbb{S}^1)=Q(\mathbb{S}^1)$ (which are essential spectra of $P(T)$, $Q(T)$ as noted in the comments) it follows that $P=f(z^n)$, $Q=f(wz^m)$ for some number $w$, $|w|=1$, positive integers $n,m$ and polynomial $f$. Further, I claim that $n=m$, it would mean just $Q(z)=P(wz)$. Indeed, choose complex number $a$, $|a|<1$, so that $f(a)\notin P(\mathbb{S}^1)$, and consider polynomials $P(z)-f(a)$, $Q(z)-f(a)$. They do not have roots on $\mathbb{S}^1$, thus the numbers of their roots inside the unit disc must be the same (as noted in your post, these numbers are just Fredholm indices of $P(T)-f(a)$, $Q(t)-f(a)$). But these numbers equal $n\cdot k$, $m\cdot k$ resp., where $k>0$ is the number of roots of $f(z)-f(a)$ in the open unit disc.  Thus $n=m$ as desired.

  [1]: http://mathoverflow.net/a/229154/4312