This is not true. Indeed, suppose that $X_k=X_{s;k}=k+sZ_k$, where $s\downarrow0$ and the $Z_k$'s are any iid random variables (r.v.'s).
To obtain a contradiction, suppose that, for the random Borel measure $\mu_s$ over $\mathbb R$ defined by $\mu_s(B):=\sum_{k\in\mathbb Z}1(X_{s;k}\in B)$, the distribution of the random variable (r.v.) $\mu_s(B)$ is Poisson with parameter $\lambda(s)|B|$ for some $\lambda(s)>0$ and all Borel $B$, where $|B|$ is the Lebesgue measure of $B$.
Note that
\begin{equation}
\mu_s((-1/2,1/2))\to1 \tag{1}
\end{equation}
in probability (see details on (1) below). Therefore and because the r.v. $\mu_s((-1/2,1/2))$ has the Poisson distribution with parameter $\lambda(s)$,
necessarily $\lambda(s)\to0$ and hence $\mu_s((-1/2,3/2))\to1$ in probability. However, similarly to (1) we have $\mu_s((-1/2,3/2))\to2$ in probability, a contradiction.
So, the random measure $\mu_s$ cannot be Poisson for all $s>0$.
Proof of (1): Note that $\mu_s(B)=\sum_{k\in\mathbb Z}1(Z_k\in\frac{B-k}s)$ and hence
\begin{equation}
1-\mu_s((-1/2,1/2))=s_1-s_2,
\end{equation}
where
\begin{equation}
s_1:=1-1\Big(Z_0\in\Big(\frac{-1/2}s,\frac{1/2}s\Big)\Big),
\end{equation}
and
\begin{equation}
s_2:=\sum_{k\in\mathbb Z\setminus\{0\}}1\Big(Z_k\in\Big(\frac{-1/2-k}s,\frac{1/2-k}s\Big)\Big).
\end{equation}
Next,
\begin{equation}
Es_1=1-P\Big(Z_0\in\Big(\frac{-1/2}s,\frac{1/2}s\Big)\Big)\to0
\end{equation}
and
\begin{equation}
Es_2=\sum_{k\in\mathbb Z\setminus\{0\}}P\Big(Z_0\in\Big(\frac{-1/2-k}s,\frac{1/2-k}s\Big)\Big)\le Es_1.
\end{equation}
Therefore and because $s_1,s_2\ge0$, we have
\begin{equation}
E|\mu_s((-1/2,1/2))-1|\le Es_1+Es_2\to0.
\end{equation}
So, by Markov's inequality, (1) follows.
$\newcommand\la\lambda\newcommand\de\delta\newcommand\R{\mathbb R}$This maximum (or, rather, supremum) distance, say $M$, is $\infty$ almost surely (a.s.).
Indeed, recall that a (simple) Poisson point process of intensity $\la\in(0,\infty)$ on $\R^d$ is a random Borel measure $m$ over $\R^d$ such that, for any pairwise disjoint bounded Borel subsets $A_1,\dots,A_k$ of $\R^d$, the random variables (r.v.'s) $m(A_1),\dots,m(A_k)$ are independent Poisson r.v.'s with respective parameters $\la|A_1|,\dots,\la|A_k|$, where $|\cdot|$ is the Lebesgue measure. It is not hard to show (see e.g. Proposition 9.1.III (ii, iii), p. 4) that $m=\sum_{i=1}^\infty\de_{X_i}$, where $\de_x$ is the Dirac delta measure supported on $\{x\}$, for $x\in\R^d$, and
the $X_i$'s are random points in $\R^d$ that are a.s. pairwise distinct. So,
$$M=\sup_i\min\{\|X_k-X_i\|\colon k\ne i\}$$
a.s., where $\|\cdot\|$ is the Euclidean norm.
Now take any real $a>0$ and any natural $n$. Take the hypercube $C_{a,n}:=[0,3na)^d$ and partition it naturally into $n^d$ congruent smaller hypercubes each with edgelengths $3a$. In each of these $n^d$ hypercubes $C_j$ ($j=1,\dots,n^d$) each with edgelengths $3a$, take the central sub-hypercube, say $B_j$, with edgelengths $a$.
Note that
$$p:=P\big(m(B_j)=1,m(C_j\setminus B_j)=0\big) \\
=P\big(m(B_1)=1\big)\,P\big(m(C_1\setminus B_1)=0\big)>0$$
for each $j=1,\dots,n^d$.
Then the probability $P(M\ge a)$ will be no less than the probability that at least one of the $n^d$ congruent smaller hypercubes $C_j$ contains exactly one point of the random points $X_i$ and that one point is in $B_j$. The latter probability is
$$1-(1-p)^{n^d}\to1$$
as $n\to\infty$. So, $P(M\ge a)\ge1$ for all real $a>0$, and hence $P(M=\infty)=1$.
Best Answer
This figure (copied from these notes) serves to illustrate the difference between a binomial and a Poisson point process. Shown are $n=100$ points randomly and independently placed in the unit square. This is a binomial point process, because the distribution of the number of points in an area $W$ inside the unit square is the binomial distribution with $n$ attempts and success probability $p=W$.
The same figure can also be seen as a realization of Poisson point process, given that there are $n=100$ points in the unit square. If this is not given, different realizations of the Poisson process will have different $n$'s, and the distribution of $n$ is the Poisson distribution. This is the only difference between the two processes, the binomial point process has a fixed number of points, while in the Poisson point process this number varies between different realizations.
If $n$ is large enough, the number of points in the unit square for the Poisson process will be peaked around the average of the Poisson distribution, so you might neglect the fluctuations around the average and just fix $n$ at the average value. In that approximation the Poisson and binomial point processes amount to the same thing.