Let $\tau_1, \tau_2,\dots,\tau_{n-1}$ be the elementary symmetric polynomials in $x_2,x_3,\dots,x_n$. It is straightforward to show that $\sigma_k=\tau_k + x_1 \cdot \tau_{k-1}$ for $k=1,2,\dots,n-1$ (with $\tau_0=1\,$).
It follows that each $\tau_k$ can be written as a polynomial $t_k$ in $x_1$ and $\sigma_1, \sigma_2,\dots,\sigma_{n-1}$:
$$
\begin{align}
\tau_1 &= \sigma_1 - x_1\,\tau_0 = \sigma_1 - x_1 = t_1(x_1, \sigma_1)
\\ \tau_2 &= \sigma_2 - x_1\,\tau_1= \sigma_2 - x_1\,t_1(x_1,\sigma_1) = t_2(x_1, \sigma_1, \sigma_2)
\\ \dots
\\ \tau_{n-1} &= \sigma_{n-1} - x_1 \, \tau_{n-2} = \sigma_{n-1} - x_1 \, t_{n-2}(x_1, \sigma_1,\dots,\sigma_{n-2}) = t_{n-1}(x_1, \sigma_1,\dots,\sigma_{n-1})
\end{align}
$$
A polynomial that is symmetric under permutations of the $x_i$’s that fix $x_1$ can then be written as:
$$
\begin{align}
f(x) &= \sum_{j=0}^n \;p_j(\tau_1, \tau_2, \dots,\tau_{n-1})\,x_1^j
\\ &= \sum_{j=0}^n \;p_j\left(t_1(x_1, \sigma_1), t_2(x_1, \sigma_1,\sigma_2), \dots,t_{n-1}(x_1, \sigma_1,\sigma_2,\dots,\sigma_{n-1})\right)\,x_1^j
\\ &= p(x_1, \sigma_1,\sigma_2,\dots,\sigma_{n-1})
\end{align}
$$
This proves the proposition for polynomials, and the extension to rational functions follows.
There is no such bound.
Given an equation $X$ over $\mathbf{Q}$ with infinitely many rational points, you can always scale the coefficients so that any finite subset of the rational points all become integral points. (Put them all under a common denominator $N$, then replace each variable $x$ by $x/N$.) Take an elliptic curve $E$ with positive rank, that is, $E(\mathbf{Q})$ is infinite. (They exist: for example, $E: y^2 = x^3 - 2$.) After scaling, you can find an elliptic curve with as many integral points as you like. (You can take $y^2 = x^3 - 2 N^6$ for the greatest common denominator $N$ of the finite set of rational points.) If such a bound in your question existed, then, since these curves all have the same degree, at some point these scaled elliptic curve would have to have infinitely many integral points. But $E(\mathbf{Z})$ is always finite for any elliptic curve by a theorem of Siegel.
Best Answer
I think doing it with mathematical induction on the number of variables $n$ is a good idea. Also, rationality of the coeffecients isn't necessary, maybe there is another way that explicitely exploits that, but the statement is true assuming just real or even complex coefficients.
The inductions start with $n=1$ is easy, a non-zero polynomial of one variable of degree $m$ can have at most $m$ different roots, and since there are infinitely many integers, the polynomial must be identical to $0$.
So how to do the missing (from OP's post) induction step ($n-1 \to n$)?
We take $x_n$ and write $p$ as a polynomial in $x_n$, where the coefficients aren't constants but polynomials in the remaining $n-1$ variables:
$$p(x_1,\ldots,x_n) = p_m(x_1,\ldots,x_{n-1})x_n^m + p_{m-1}(x_1,\ldots,x_{n-1})x_n^{m-1}+\ldots + p_2(x_1,\ldots,x_{n-1})x_n^2 + p_1(x_1,\ldots,x_{n-1})x_n + p_0(x_1,\ldots,x_{n-1}) = \sum_{i=0}^m p_i(x_1,\ldots,x_{n-1})x_n^i,$$
where $m$ is the highest degree that $x_n$ occurs in $p$.
Note that this works even if $p$ doesn't really depend on $x_n$, then $m=0$ and the following arguments still works on $p_0$ alone.
If all of the $p_i(x_1,\ldots,x_{n-1}), i=0,1,\ldots,m$ are the zero polynomial, we are done. If not, then we have a $k$ such that $p_k(x_1,\ldots,x_{n-1})$ is not identical to $0$. From the induction hypothesis we now know that there exist integers $x_1^*, \ldots, x_{n-1}^*$ such that $p_k(x_1^*,\ldots,x_{n-1}^*) \neq 0$.
That means the polynomial
$$q(x_n)=p(x_1^*,\ldots,x_{n-1}^*,x_n) = \sum_{i=0}^m p_i(x_1^*,\ldots,x_{n-1}^*)x_n^i$$
is not identical to $0$, at least the term $p_k(x_1^*,\ldots,x_{n-1}^*)x_n^k$ is not zero. But know we are back at the induction start $n=1$: We have a non-zero polynomial $q$ in one variable, so there must be some integer $x_n^*$ such that $q(x_n) \neq 0$ and that finally means $p(x_1^*,\ldots,x_{n-1}^*,x_n^*) \neq 0$.
That concludes the induction step, we have proved that any non-zero polynomial in $n$ variables cannot have roots at every point in $\mathbb Z^n$.