Here is an argument avoiding Siegel's theorem (and also seemingly different from the DLS argument to which Dimitrov refers, as the main tool I will use is Weil's RH for absolutely irreducible curves whereas DLS use arguments based on Hilbert irreducibility).

Let $S$ be a non-empty finite set of primes (e.g., the primes factors of $d$) and consider $f \in \mathbf{Z}[X]$ that is not an $e$th power in $\mathbf{Z}[X]$ (equivalently, in $\mathbf{\mathbf{Q}}[X]$) for each $e \in S$. An elementary argument with the monic multiple of $f$ shows that for each of the primes $e \in S$ the polynomial $f$ either (i) is not an $e$th power in $\overline{\mathbf{Q}}[X]$ or (ii) is of the form $c h^e$
for some monic $h$ with $c$ the leading coefficient of $f$.
If there is any integer $n$ away from zeros of $f$ such that $f(n)$ is an $e$th power for some $e$ as in case (ii) then $c$ is an $e$th power and hence so is $f$. Thus, we can assume for our purposes that case (ii) never occurs.

Since each $e\in S$ is *prime*, it follows that the polynomial $Y^e - f(X)$ is irreducible in $\overline{\mathbf{Q}}[X,Y]$ (as it is the same to be irreducible in $\overline{\mathbf{Q}}(X)[Y]$, for which $f$ not being an $e$th power in $\overline{\mathbf{Q}}(X)$ is equivalent to the irreducibility property since $e$ is prime).
We conclude that $Y^e - f(X)$ is absolutely irreducible over $\mathbf{Q}$ for every $e \in S$. Hence, for all large primes $p$, $Y^e - f(X)$ is absolutely irreducible over $\mathbf{F}_p$ too. (Recall that "absolute irreducibility" is inherited under reduction modulo all but finitely many primes, whereas ordinary irreducibility is not.) In what follows, only consider such $p$ (moreover big enough so that $f \bmod p$ has the same degree as $f$).

These curves $Y^e - f(X) = 0$ for varying $e \in S$ (if $\#S > 1$) have *finite* overlap in characteristic 0, so they are pairwise disjoint up to uniformly bounded error in characteristic $p$ for large $p$. Hence, the solution set $V_p$ to $h \equiv 0 \bmod p$ is the "disjoint" (up to bounded amount) union of the solutions sets to the individual curves $C_{e,p} := \{y^e - f(x) = 0\}$. There are at most $d := \deg(f)$ values $x \in \mathbf{F}_p$ where $f$ vanishes mod $p$, over which there is only one point $(x,0)$ in $V_p$. Ignoring those at most $d$ points, as well as the uniformly bounded overlaps sets for distinct $e$'s just mentioned, every other fiber of $V_p$ over the $x$-line $\mathbf{F}_p$ lies in exactly one of the curves $C_{e,p}$.

Consider $p \equiv 1 \bmod e$ for all $e \in S$. The fibers for $C_{e,p}$ have size $e$ (away from zeros of $f$ in $\mathbf{F}_p$). As $p$ grows, $\#C_{e,p}(\mathbf{F}_p) \sim p$ for each $e \in S$, by RH, so the image of $C_{e,p}$ in $\mathbf{F}_p$ consists of $\sim p/e$ points as $p$ grows.

Varying through all $e \in S$, if $V_p$ actually hits the *entire* $x$-line for all such large $p$ (say even up to a bounded amount as such $p$ grows) we would get
$\sum_{e \in S} 1/e = 1$ (equality on the nose, not just approximation). But the $e$'s are pairwise distinct primes, so no such equality is possible (look at it $e_0$-adically for one $e_0 \in S$).

Thus, for (many) large $p$ the projection $x: V_p \rightarrow \mathbf{F}_p$ is not surjective, so if $n \in \mathbf{Z}$ represents a mod-$p$ residue class not in the image then for every $e \in S$ the congruence $y^e \equiv f(n) \bmod p$ has no solution, so certainly for every $e \in S$ the integer $f(n)$ is not an $e$th power in $\mathbf{Z}$ as well.

QED

Polynomials of certain special type, Acta Arithmetica IX, 1964 - or Schinzel's collected works. $\endgroup$ – Vesselin Dimitrov Nov 30 '15 at 3:44