Let's Recall the statement of the theorem:
Theorem: (Rouché) Let $A$ be some region enclosed by a simple loop $\gamma$. If $f(z)$ and $g(z)$ are analytic on an open neighborhood of $A$, and $|f(z)|>|g(z)|$ on $\gamma$, then $f(z)$ and $f(z)+g(z)$ have the same number of zeros inside $A$.
Let $p(z)=z^4+z^3-4z+1$ be your polynomial. First lets count the number of zeros inside the disk $|z|=1$. Since $$|-4z|\geq 3\geq |z^4+z^3+1|,$$ we know that there will be one zero in this region by applying Rouché's Theorem with $f=-4z$.
Circle |z|=2: To find the number of zeros inside the circle $|z|=2$, we have to be slightly trickier. Notice that we have the miraculous identity
With some manipulation of inequalities we also have the following lemma
Lemma: The inequality $$2|z-1||z+1|^2> |z+1|^2+2$$ holds for all $|z|=2$.
Consider $f(z)=z(z-1)(z+1)^2$ and $g(z)=(z+1)^2-z$ in Rouché's Theorem. When $|z|=2$, we have both $$|(z+1)^2|+2\geq |(z+1)^2-z|,$$ and $$|z(z-1)(z+1)^2|\geq 2|(z-1)(z+1)^2|.$$ Hence by the lemma $|f(z)|>|g(z)|$ for $|z|=2$ so that $p(z)$ has four roots inside the disk $|z|=2$.
Conclusion: As there are no zeros with $|z|=2$ or $|z|=1$, we conclude there are $3$ zeros in the region $1<|z|<2$.
Hope that helps,