How can I guarantee the unique positive root of this polynomial?

Descartes's sign rule ensures that both polynomials have exactly $1$ positive root for every $n$.

Consider $f(x)=x^{n+1}+ x^n -1.$ Note that $f(0)=-1.$ Now, since $\lim_{x\to \infty} f(x)=+\infty$ there exists $a>0$ such that $f(a)>0.$ Since $f$ is continuous in $[0,a]$ it follows from Bolzano's theorem that there is a root in $(0,a).$ A similar argument works with $f(x)=x^{n+1}- x^n -1.$

Now, if $f(x)=x^{n+1}+ x^n -1$ then $f'(x)=(n+1)x^n+nx^{n-1}>0$ on $(0,\infty).$ Thus, $f$ is strictly increasing, from where it has at most one root. Since it has one, it has exactly one.

Now, if $f(x)=x^{n+1}- x^n -1$ then $f(x)<0$ in $(0,1).$ And $f'(x)=(n+1)x^n-nx^{n-1}>0$ on $(1,\infty).$ Thus, $f$ is strictly increasing in $(1,\infty),$ from where it has at most one root. Since it has one, it has exactly one.

The derivative of this polynomial is

$$(n + 1) x^n + n x^{n - 1} = x^{n - 1} \big((n + 1) x + n\big)$$

This has zeros at $x = 0$ and $x = -\frac{n}{n + 1} < 0$; in particular, it is positive for $x > 0$. Thus the original polynomial is strictly increasing from the point $(0, -1)$, and has a unique positive root.