According to Fundamental theorem of algebra it must have 5 zeros in some region.
If we chose
$f(z) = z^5$ and
$g(z)= -3z^2 -1$
Then $|f(z)| > |g(z)|$
This condition is satisfied for the positive integers such that $|z|≥2$.
We chose $|z| = 2$
Then according to Rouché's theorem
$f(z)$ and $f(z)+g(z)$ will have same number of zeros inside $|z|=2$
It means that all the five zeros will lie inside $|z| = 2$
And there will be no zeros outside for $|z| > 2$
This was my solution
But according its correct ans is $|z|>2^⅔$
I did not get.
How this could be get.
Observe that $$ (2^{2/3})^5-3\cdot(2^{2/3})^2-1=8\cdot 2^{1/3}-6\cdot 2^{1/3}-1=2\cdot 2^{2/3}-1>0. $$ Thus $|z|^5>3|z^2|+1\ge |3z^2-1|$ on the circle $|z|=2^{2/3}$ and hence $c=2^{2/3}$ works.
You can improve it a little bit, if you replace $c=2^{2/3}$ with $c=1.51<2^{2/3}$, and the same argument works.
In fact, the optimal $c$ is a root of the equation $z^5-3z^2-1=0$, which is equal to $$ c=1.50941045653627123243833773286186\ldots $$