Parse the minimum value of the required $m$ as $\left\lfloor\frac{k}{2}\right\rfloor!\left\lceil\frac{k}{2}\right\rceil!+1$ (where $\lceil x\rceil$ and $\lfloor x\rfloor$ represent the smallest integer not less than $x$ and the largest integer not greater than $x$, respectively).

First, prove: $m \geqslant\left\lfloor\frac{k}{2}\right\rfloor!\left\lceil\frac{k}{2}\right\rceil!+1$.
In fact, suppose there exists a function $f(x)$ that satisfies the condition, and let $x_{1}, \cdots, x_{k}$ be the $k$ distinct integer roots of $f(x)-m$. We can write $f(x)-m=\left(x-x_{1}\right) \cdots\left(x-x_{k}\right) q(x), q(x) \in \mathbf{Z}[x]$, thus
$$m=(f(x)-1)+1-\left(x-x_{1}\right) \cdots\left(x-x_{k}\right) q(x) .$$

Since $f(x)-1$ has an integer root, let $\alpha$ be its integer root, then
$$m=1-\left(\alpha-x_{1}\right) \cdots\left(\alpha-x_{k}\right) q(\alpha) .$$

By $m>1$, we know $-\left(\alpha-x_{1}\right) \cdots\left(\alpha-x_{k}\right) q(\alpha)>0$, so the numbers $\alpha-x_{1}, \cdots, \alpha-x_{k}$ are $k$ distinct non-zero integers, $\left.\left|\left(\alpha-x_{1}\right) \cdots\left(\alpha-x_{k}\right)\right| \geqslant\left\lfloor\frac{k}{2}\right\rfloor!\left\lceil\frac{k}{2}\right\rceil!\right.$. From (1), we know $\left.m \geqslant\left\lfloor\frac{k}{2}\right\rfloor!\left\lceil\frac{k}{2}\right\rceil!+1\right.$.

Next, prove: When $m=\left\lceil\frac{k}{2}\right\rceil!\left\lfloor\frac{k}{2}\right\rfloor!+1$, there exists a polynomial that meets the requirements.
When $k$ is even, let $k=2 n$, then $m=(n!)^{2}+1$, let $f(x)=(-1)^{n-1}(x+1) \cdots (x+n)(x-1) \cdots(x-n)+m$;

When $k$ is odd, let $k=2 n+1$, then $m=n!(n+1)!+1$, let $f(x)=(-1)^{n}(x+1) \cdots(x+n)(x-1) \cdots(x-(n+1))+m$.