Solution:

For $k=1$ the statement is trivial. So let $k>2$. Call an interval of length $k$ a set of the form $\{x, x+1, \ldots, x+k\}, x \in \mathbb{Z}$. Two integers $x$ and $y$ will be called consecutive if and only if there exist intervals $I_{1}$ and $I_{2}$ of length $k$ for which $I_{1} \cap I_{2}=\{x, y\}$. However, by the condition of the problem $f\left(I_{1}\right)$ and $f\left(I_{2}\right)$ are also intervals of length $k$, so since $\{f(x), f(y)\}=f\left(I_{1}\right) \cap f\left(I_{2}\right)$, it follows that $f(x)$ and $f(y)$ are also consecutive numbers. From this, $|f(x+1)-f(x)|=1$ for $x \in \mathbb{Z}$. Finally, using the injectivity of the mapping, by a simple induction on $n$ we obtain that $|f(x+n)-f(x)|=n$.