Since $\deg f > \deg g$, when $x$ is sufficiently large we have $|g(x)/f(x)| < 1$. That is, there exists a real number $R$ such that for all $|x| > R$, we have $|g(x)/f(x)| < 1$. For all such $x$ and all primes $p$, we have
$$|pf(x) + g(x)| \geq |f(x)| \left( p - \frac{|g(x)|}{|f(x)|} \right) > 0.$$
Therefore all real roots of the polynomial $pf + g$ lie in $[-R, R]$.

Let $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$ and $g(x) = b_m x^m + b_{m-1} x^{m-1} + \cdots + b_0$, where $n > m$, $a_n \neq 0$ and $b_m \neq 0$. Replacing $f(x)$ and $g(x)$ by $a_n^{-1} f(x/a_n)$ and $a_m^{-1} g(x/a_m)$ respectively, we reduce the problem to the case $a_n = 1$, in which the leading coefficient of $pf + g$ is $p$. If $r = u/v$, $(u, v) = 1$ and $v > 0$, is a rational root of $pf + g$, then $v$ is either $1$ or $p$.

Suppose there are infinitely many instances with $v = 1$. If $v = 1$ then $|u| \leq R$, so there are only finitely many integers $u$. Hence there exist distinct primes $p$ and $q$ such that we get the same value of $u$. Then the polynomials $pf + g$ and $qf + g$ share a common root, from which we obtain $f(u) = g(u) = 0$. So in this case $f$ and $g$ have a common integer root.

Suppose there are infinitely many instances with $v = p$. Comparing the powers of $p$ in the denominators of $pf(u/p)$ and $g(u/p)$, we take $m = n - 1$, so that $pf(u/p) + g(u/p) = 0$ simplifies to
$$(u^n + a_{n-1} p u^{n-1} + \cdots + a_0 p^n) + (b_{n-1} u^{n-1} + b_{n-2} p u^{n-2} + \cdots + b_0 p^{n-1}) = 0.$$
This shows that $u^n + b_{n-1}u^{n-1}$ is divisible by $p$, and since $(u, p) = 1$, we get $u + b_{n-1} = pk$, where $k$ is some integer. Also, all roots of $pf + g$ lie in $[-R, R]$, so
$$\frac{|pk - b_{n-1}|}{p} = \frac{|u|}{p} < R, \quad |k| < R + \frac{|b_{n-1}|}{p} < R + |b_{n-1}|.$$
Hence there are only finitely many possible values of the integer $k$. So there exists an integer $k$ such that for infinitely many primes $p$, $\frac{pk-b_{n-1}}{p} = k - \frac{b_{n-1}}{p}$ is a root of $pf + g$. For these primes, we have
$$f\left(k - \frac{b_{n-1}}{p}\right) + \frac{1}{p}g\left(k - \frac{b_{n-1}}{p}\right) = 0.$$
Hence
$$f(k - b_{n-1}x) + xg(k - b_{n-1}x) = 0 \quad (1)$$
has infinitely many solutions of the form $x = 1/p$. Since the left-hand side of equation (1) is a polynomial, equation (1) holds for all real numbers $x$. Substituting $x = 0$ into equation (1), we obtain $f(k) = 0$. Thus the integer $k$ is a root of $f$.