Let $y=x^{1/n}$. Then $x^{n} + x^{1/n} = y^{n^2} + y$.
In particular, for any constant $k$, you have $\frac{d}{dy}\left(y^{n^2}+ y + k\right) = n^2y^{n^2-1} + 1.$
For nonnegative $y$, this is always positive; so $x^n + x^{1/n}+a$ is strictly increasing on $[0,\infty)$. (Well, technically, $y^{n^2}+y+a$ is strictly increasing; but since $y$ is itself an increasing function of $x$ for $x\geq 0$ and $n\gt0$, then so is $x^n + x^{1/n}+a$). In particular, it has at most one zero on $[0,\infty)$.
If $a\gt 0$, then the function is positive at $0$, hence has no zeros on $[0,\infty)$. If $a\leq 0$, then the function is nonpositive at $0$, hence has exactly one zero on $[0,\infty)$.