Let be $f$ analytic in the disc with radius $$1 from the beginning, and continuous on the boundary of the disc. let $f(0)=1$ and also let $z$ on the boundary satisfies $1<|f(z)|<M$. Then there can't be $|z_{0}|<\frac{1}{M}$ satisfying $f(z_0)=0$ . I thought about using Cauchy integral formula but it doesn't seem to be useful.
Edit: I thought about that idea: $M$ is the maximum value on the disc. So for every sub disc the maximum value must be $M$. Let be $z_0$ a zero on the sub-disc, then there is an environment that is smaller than $1$. Therefore the integral on the sub disc must be less than $1$, which contradicts the value in the center of the disc.
If $|f|$ is bounded by $M$ in the unit disk then $g(z) = f(z)/M$ maps the unit disk into itself.
Now let $z_0$ be a zero of $f$ in the unit disk. Then $g(0) = 1/M$ and $g(z_0) = 0$ and the Schwarz-Pick theorem shows that $$ \frac 1M = |g(0)| = \left|{\frac {g(0)-g(z_{0})}{1-\overline {g(0)}g(z_{0})}}\right|\leq \left|{\frac {0-z_0}{1-\overline {0}z_0}}\right| = |z_0| \, . $$ So every zero $z_0$ of $f$ in the unit disk satisfies $|z_0| \ge 1/M$.