To expand on a comment of Lucia, when $c$ is irrational, we can show that there are at most $O((\log N)^2)$ values of $n\leq N$ such that $N^c$ is rational, let alone an integer.

Let $\mathcal{A}$ be the set of $n\leq N$ such that $n^c$ is rational and $n > 1$. Let $n_1$ be the smallest element of $\mathcal{A}$ and let $n_2$ be the smallest element of $\mathcal{A}$ that is not a rational power of $n_1$. The number of elements of $\mathcal{A}$ when no such $n_2$ exists is $O((\log N)^{1+\varepsilon})$ (exercise to the reader).

Since $n_2$ is not a rational power of $n_1$, given a prime $p\mid n_1n_2$, there exists another prime $q\mid n_1n_2$ such that
$$
\tag{1}\label{1}
\nu_p(n_1)\nu_q(n_2) - \nu_p(n_2)\nu_q(n_1) \neq 0,
$$
where $\nu_p(m)$ is the exponent of $p$ in the prime factorization of $m$.

Let $n_3$ be a third element of $\mathcal{A}$ that is not a rational power of $n_1$ or $n_2$ (we're done if this element doesn't exist). Then since $1$ and $c$ are linearly independent over $\mathbb{Q}$, and since
$$
\begin{gathered}
\exp\left(\log n_1\right),\ \exp\left(\log n_2\right),\ \exp\left(\log n_3\right),\\
\exp\left(c\log n_1\right),\ \exp\left(c\log n_2\right),\ \exp\left(c\log n_3\right)
\end{gathered}
$$
are all rational, the six exponentials theorem implies that the three real numbers $\log(n_1),\log(n_2),\log(n_3)$ are linearly dependent over $\mathbb{Q}$. That is, there exist rational numbers $A,B,C$ not all zero such that
$$
A\log(n_1) + B\log(n_2) + C\log(n_3) = 0.
$$
Actually, none of $A,B,C$ are zero, since this would imply that two of $n_1,n_2,n_3$ are rational powers of one another. Dividing by $C$ and renaming the coefficients, and exponentiating, we have
$$
\tag{2}\label{2}
n_3 = n_1^A n_2^B
$$
for some nonzero rational $A,B$. Since $n_3 > 1$, we must have $(n_3,n_1n_2) > 1$. Let $p\mid (n_3,n_1n_2)$ and let $q$ be given by \eqref{1}. Then
$$
\begin{aligned}
\nu_p(n_3) &= A\nu_p(n_1) + B\nu_p(n_2), \\
\nu_q(n_3) &= A\nu_q(n_1) + B\nu_q(n_2).
\end{aligned}
$$
There are $O((\log N)^2)$ possible values for the pair $(\nu_p(n_3),\nu_q(n_3))$ with $\nu_p(n_3)\neq (0,0)$. For each pair, \eqref{1} implies that this system has a unique solution in $A$ and $B$. Since any element of $\mathcal{A}$ corresponds to a choice of $A$ and $B$ in \eqref{2}, and since any solution to \eqref{2} corresponds to at most one solution to the system above, the set $\mathcal{A}$ contains at most $O((\log N)^2)$ elements.

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