Does there exist any non-zero polynomial $f:\mathbb C \to \mathbb C$ such that $f(x+2)-2f(x+1)=f(x) , \forall x \in \mathbb C$?

No polynomial can satisfy a linear recurrence except in trivial cases because the solution of linear recurrences is a combination of geometric progressions.

For instance, $f(n+2)-2f(n+1)=f(n)$ for all $n \in \mathbb N$ implies that $$f(n) = c_1 (1-\sqrt2)^n+c_2 (1+\sqrt2)^n$$ where $c_1, c_2$ are parameters determined by $f(0)$ and $f(1)$.

Indeed, $f(n) \sim c_2 (1+\sqrt2)^n$ as $n \to \infty$, but a polynomial grows as $\sim a n^d$.

The equation says

$\Delta^2 f(x) = 2f(x)$

where $\Delta=\Delta_{+1}$ is the finite difference operator $\Delta f = f(x+1) - f(x)$.

Thus $\Delta^k$ does not kill $f$ no matter how large $k$ is, so $f$ cannot be a polynomial.

General case. A linear recursion on $f(x)$ is consistent with it being a polynomial if and only if it is equal to the result of taking some linear relation satisfied by the $0$ function, and applying a finite sequence of difference operators $\Delta_a \circ \Delta_b \circ ...$ to that. This does not assume the shifts $a,b,...$ are integers.