With the conditions as stated, often times there will be no root in the range you give. Since the derivative is $nx^{n-1}-k$, the only critical point occurs at $$b=\left(\frac{k}{n}\right)^{1/(n-1)}.$$ As $1$ is always a root and there is only one other root to the other side of $b$, whenever $k\leq n$ the only other root will be less than $1$.
For instance, to get a root in the range you want the root of the derivative should be between $1$ and $1.15$, imposing the constraint
$$
\left(\frac{k}{n}\right)^{1/(n-1)}\leq1.15;
$$
that is, you require
$$
\bbox[5px, border: 2px solid red]{k\leq n\,1.15^{n-1}.}
$$
Note that this is a necessary condition, but not sufficient, for a root to be in your range.
As for finding said root, Newton's method works very fast, with the recursion
$$
\bbox[5pt, border: 2pt solid green]{x_{m+1}=x_m-\frac{x_m^n-kx_n+k-1}{nx_m^{n-1}-k}.}
$$
Newton's method has the advantage that most often you can start really far from the root and still get there.
Caveat: in this case, with only two roots, the choice of the starting point of the method is easy. In general, though, it could be really complicated, as this example shows.