Solution:
Assume that we have a polynomial $P$ that satisfies the desired conditions. We begin by examining some easy cases:
If $P$ is a constant polynomial, that is, $P(z)=c$ for some $c \in \mathbb{R}$, the condition becomes $c z=1$ if and only if $c^{2}=0$. If $c=0$, then $c^{2}=0$ is always satisfied while $c z=1$ cannot hold, contradiction. If however $c \neq 0$, we can plug in $z=\frac{1}{c}$ to obtain a contradiction. Therefore, there are no solutions among constant polynomials.

What about polynomials of the form $P(z)=m z+b$ where $m \neq 0$? The condition becomes $m z^{2}+b z=1$ if and only if $(m(z-1)+b)(m(z+1)+b)=0$, so the two quadratic polynomials $m z^{2}+b z-1$ and $m^{2} z^{2}+2 b m z+\left(b^{2}-m^{2}\right)$ have the same roots, and this means that they are multiples of each other. By comparing the leading coefficients we see that the multiplying factor is $m$, and by comparing the other coefficients we get that $b m=2 b m$ and $-m=b^{2}-m^{2}$. Since $m \neq 0$, the first of these two equations gives us $b=0$ and the second $m=1$. This shows that the only linear polynomial that can satisfy these conditions is $P(z)=z$, which is obviously a solution.

Let's now assume that $\deg(P)=n \geq 2$.
Consider the polynomial $Q(z)=(z+1) P(z+1)-(z-1) P(z-1)$. It's easy to see that $\deg(Q) \leq \deg(P)$ (since the highest degree terms of $(z+1) P(z+1)$ and $(z-1) P(z-1)$ cancel out). We show that $Q$ also has the same roots as $P$:
If $P(r)=0$ for some complex number $r$, by plugging in $z=r \pm 1$ in the condition of the problem, we see that $(r-1) P(r-1)=1=(r+1) P(r+1)$ and therefore $Q(r)=0$. This shows that all roots of $P$ are also roots of $Q$. Since $P$ has no repeated roots and the degree of $Q$ is not greater than the degree of $P$, we deduce that $Q$ is just a multiple of $P$.

Let $P(z)=\sum_{k=0}^{n} a_{k} z^{k}$. This gives us
$$Q(z)=\sum_{k=0}^{n} a_{k}\left((z+1)^{k+1}-(z-1)^{k+1}\right)$$
In order to compare $Q$ to $P$, we would like to write it in the form $Q(z)=\sum_{k=0}^{n} b_{k} z^{k}$. To compute $b_{n}$, note that the only terms in $(\star)$ that contain $z^{n}$ come from $k=n-1$ and $k=n$. We get
$$b_{n} z^{n}=a_{n-1}\left(z^{n}-z^{n}\right)+a_{n}\left((n+1) z^{n}-\left(-(n+1) z^{n}\right)\right)=2(n+1) a_{n} z^{n}$$
so we must have $Q=2(n+1) P$. Now we compute $b_{n-1}$. Similarly to above, we see that the only relevant terms in $(\star)$ come from $k=n-2, k=n-1$ or $k=n$. We get
$$\begin{aligned} b_{n-1} z^{n-1} & =a_{n-2}\left(z^{n-1}-z^{n-1}\right)+a_{n-1}\left(n z^{n-1}-(-n) z^{n-1}\right)+a_{n}\left(\binom{n+1}{2} z^{n-1}-\binom{n+1}{2} z^{n-1}\right) \\ & =2 n a_{n-1} z^{n-1} \end{aligned}$$
But if $Q=2(n+1) P$, we must have $b_{n-1}=2(n+1) a_{n-1}$, contradiction! This shows that there cannot be any polynomial of degree $\geq 2$ satisfying the desired conditions.

We give a different argument for the case $n \geq 2$:
Note that the problem statement is equivalent to the statement that the two polynomials $z P(z)-1$ and $P(z-1) P(z+1)$ have the same set of roots. However, the polynomial $z P(z)-1$ has degree $n+1$ and thus at most $n+1$ different roots. On the other hand, for each of the distinct roots $r_{1}, r_{2}, \ldots, r_{n}$ of $P$, the numbers $r_{1}+1, \ldots, r_{n}+1$ are roots of $P(z-1)$ and the numbers $r_{1}-1, \ldots, r_{n}-1$ are roots of $P(z+1)$. Write $r_{k}=x_{k}+y_{k} i$. For any fixed $y$ we note the following:
If there are $m$ roots of $P$ with imaginary part $y$, then there are at least $m+1$ distinct numbers among $\left\{r_{1}-1, \ldots, r_{n}-1, r_{1}+1, \ldots, r_{n}+1\right\}$ with imaginary part $y$. This is because if WLOG $r_{1}, \ldots r_{m}$ all have imaginary part $y$ and real parts $x_{1}<\ldots<x_{m}$ (the strict inequality holds because we cannot have repeated roots), then the $m+1$ numbers $r_{1}-1, r_{1}+1, \ldots, r_{m}+1$ all have different real part and so they must be different.
Since we cannot have more than $n+1$ roots, we conclude that $y_{1}=\ldots=y_{n}$. In fact, they are all equal to zero because we know that if $x_{k}+y_{k} i$ is a root of a polynomial with real coefficients, then so is $x_{k}-y_{k} i$. But since all $y_{k}$ are equal, we must have $y_{k}=-y_{k}$.
Since $y_{1}=\ldots=y_{n}$, the $x_{k}=r_{k}$ are pairwise distinct and we can order them $x_{1}<\cdots<x_{n}$. Analogously to above, we have $x_{1}-1<x_{1}+1<\cdots<x_{n}+1$, so those have to be our $n+1$ roots. Clearly, $x_{k}-1$ is the $k$-th smallest root, and by comparing to our chain of inequalities above, we must have $x_{k}-1=x_{k-1}+1$, or $x_{k}=x_{k-1}+2$.
All in all, we learned that all the roots of $P(z)$ are real and form an arithmetic progression of difference 2. We can therefore write:
$$P(z)=c \cdot(z-a+2) \cdot(z-a+4) \cdot \ldots \cdot(z-a+2 n)$$
for some constants $a, c \in \mathbb{R}$ where $c \neq 0$. Hence, we can also write:
$$z P(z)-1=c \cdot z \cdot(z-a+2) \cdot \ldots \cdot(z-a+2 n)-1$$
The conditions in the problem statement now imply that the roots of the polynomial $z P(z)-1$ are $\{a-1, a-3, \ldots, a-(2 n+1)\}$. Plugging in $z=a-1$ into the equation above, we get
$$1=c \cdot(a-1) \cdot 1 \cdot 3 \cdot \ldots \cdot(2 n-1)$$
On the other hand, plugging in $z=a-(2 n+1)$ gives
$$1=c \cdot(a-(2 n+1)) \cdot(-(2 n-1)) \cdot \ldots \cdot(-3) \cdot(-1)$$
By taking absolute values of $(\star)$ and $(\star \star)$ and using $c \neq 0$, we see that $|a-1|=|a-(2 n+1)|$, so $a=n+1$, which allows us to find $c>0$ using $(\star)$. But now we can plug $z=a-3=n-2$ into the equation for $z P(z)-1$ above to obtain
$$1=c \cdot(n-2) \cdot(-1) \cdot 1 \cdot \ldots \cdot(2 n-3)$$
Since $n \geq 2$, each factor on the RHS except for the $-1$ is non-negative, which means that this equation cannot hold! We conclude that there is no such $P$ for $n \geq 2$.

Therefore, the only solution is $P(z)=z$.