It is clear by definition, that both $P_n(z)$ and $Q_n(z)$ have $n$ zeros in the complex plane and that $P_n$ and $Q_n$ cannot have the same zeros.
In addition, $P_n(z) \to e^z$ as $n\to\infty$ (uniformly) on $\overline{D(0,r_k)}$, where $r_k>0$ is some radius.
This means that for any $r_k>0$, $P_n(z)$ contains no zeros in $\overline{D(0,r_k)}$ (for $n$ large enough), hence
they must be outside the closed disc.
Now consider all the zeros of $Q_n$ that are not 0. $Q_n(z) = 0 \iff e^z-1=0$.
Also, $e^z - 1 = 0 \iff z=2\pi i k$, where $k\in\mathbb{Z}\setminus\{0\}$.
Going forward, let $k = |k|$.
Hence for $2\pi k < r_k < 2\pi (k+1)$ there exists $\epsilon_k>0$ such that
\begin{equation*}
\begin{aligned}
|e^z-1| > \epsilon_k
\end{aligned}
\end{equation*}
From uniform convergence, for all $\epsilon_k>0$ there exists $N_k\in\mathbb{N}$ such that for all $n\ge N_k$
\begin{equation*}
\begin{aligned}
|e^z - 1 - Q_n(z)| = |e^z - P_n(z)| < \epsilon_k
\end{aligned}
\end{equation*}
Let $\gamma$ be the positively oriented circle with centre $0$ and radius $r_k$ (its range in the plane is $\gamma^*$), then for all $n\ge N_k$
\begin{equation*}
\begin{aligned}
|e^z - 1 - Q_n(z)| < |e^z - 1| \quad (\forall z \in\gamma^*)
\end{aligned}
\end{equation*}
Then by Rouche's Theorem, for all $n\ge N_k$, $e^z-1$ has the same number of zeros as ${Q_n(z)}$ in $D(0,r_k)$,
which is just $2k+1$ (positive and negative integers as well as $z=0$).
Hence $Q_n(z)$ has $n - (2k+1) = n- 2k - 1$ zeros in $\mathbb{C}\setminus D(0,r_k)$.
Since $2\pi k < r_k < 2\pi (k+1)$ then $2\pi (k+1) < r_k + 2\pi = r_{k+1} < 2\pi (k+2)$.
We repeat this same process.
Hence $Q_n(z)$ has $2(k+1)+ 1 = 2k+3$ zeros in $D(0;r_{k} + 2\pi)$ and $n-(2k+3) = n-2k-3$ in $\mathbb{C}\setminus D(0;r_{k} + 2\pi)$, this is
for all $n\ge N_{k+1}$.
Hence there exists $M_k \in \mathbb{N}$ such that for all $n\ge M_k$, we have that $Q_n(z)$ has $2$ zeros in the annulus $A = \{ z\in\mathbb{C} : r_k < |z| < r_k +2\pi\}$,
$2k+1$ zeros in $D(0;r_k)$ and $n-2k-3$ zeros in $\mathbb{C}\setminus D(0;r_k+2\pi)$.
It is also clear that $Q_n(z)$ always has a zero at $z = 0$.
It has been shown in a paper by K. S. K. Iyengar that the zeros of $P_n(z)$
lie in the annulus $A = \{ z\in\mathbb{C}: \frac{n}{e^2} < |z| < n \}$ for large $n$.
It has also been shown in a paper by Buckholtz that none of the zeros of $P_n(nz)$ (rescaled) lie in $C = \{z\in\mathbb{C}: |ze^{-1/z}| \le 1 \text{ and } |z| \le 1\}$ for large $n$.
