Alexandre Eremenko already showed that for each $n>1$ the function $$ S_n(x) := \frac1{n!} \sum_{k=1}^n {n \choose k} (-1)^{n-k} k^x $$ has infinitely many zeros $x \in {\bf C}$. One can still say more: for each $n$ (including $n=2$, and for that matter $n=1$) the zeros are limited to a vertical strip $\sigma_0 < {\mathop{\rm Re}}(x) < \sigma_1$ (this is elementary: if the real part of $x$ is too positive or too negative then the $k=n$ or $k=1$ term dominates), and the number of complex conjugate pairs with $|{\mathop{\rm Im}}(x)| < T$ is asymptotic to $c_n T$ where $c_n := (2\pi)^{-1} \log n$.
The proof is similar to the standard proof of the asymptotic vertical distribution of zeros of the Riemann zeta function, but easier because $S_n$ is an elementary function. Use the argument principle to express the number of zeros in the rectangle $\sigma_0 < {\mathop{\rm Re}}(x) < \sigma_1$, $|{\mathop{\rm Im}}(x)| < T$ as a contour integral over its boundary. The integrals over the vertical $\sigma_0$ and $\sigma_1$ edges contribute $O_n(1)$ and $2 c_n T + O_n(1)$ respectively. For the horizontal edges: use the Hadamard factorization of $S_n$, take its logarithmic derivative, show that the number of zeros with $|{\mathop{\rm Im}}(x) - T| \leq 2$ is bounded, and show that the integral of $|S_n^{\phantom.\prime}/S_n|$ over the rectangle $[\sigma_0,\sigma_1] + i [T-1,T+1]$ is $O_n(1)$ and thus that we make the horizontal $S_n^{\phantom.\prime}/S_n$ integral also $O_n(1)$ by changing $T$ by at most $1$.
For $n=3$ one can be still more precise: for each $k=1,2,3,\ldots$, the horizontal strip $k c_3^{\phantom.} < {\mathop{\rm Im}}(x) < (k+1) c_3^{\phantom.}$ contains a zero $x_k$ of $S_3$, and the full set of zeros consists of these $x_k$, their complex conjugates, and the real zeros $x=1$ and $x=2$. This is obtained by applying Rouché's theorem to $6 S_3(x) = 3^x - 3\cdot 2^x + 3$ with comparison function $3^x + 3$. This $x_k$ can be approximated numerically by integrating $(2\pi i)^{-1} z (S_3^{\phantom.\prime}(z) / S_3(z)) dz$ around the boundary of this rectangle and then applying Newton's formula to get even closer. The first five complex zeros $x_1,x_2,x_3,x_4,x_5$ are approximately
-0.3397375 + 8.9137244 i,
2.8692517 + 15.2110263 i,
0.0637801 + 18.6324632 i,
-0.1248035 + 26.7730278 i, and
2.9811739 + 31.1087024 i.
Note that ${\mathop{\rm Re}}(x_5)$ is almost $3$. It so happens that $\sigma_3$ \sigma_1$ is exactly $3$ (since $3^3 = 2 \cdot 2^3 + 3$). Of the $174$ zeros with $0 < {\mathop{\rm Im}}(x) < 1000$, the one with largest real part is $x_{116} \doteq 2.99976958 + 666.32539172i$. The least real part in that range is attained by $x_{158} \doteq -0.36455251 + 906.47874219i$ (while $\sigma_0 \doteq -0.3646005647$).
