The pairs $(p,q)$ satisfying the given equation are those of the form $p(x) = ax+b, q(x) = cx+d$
for $a,b,c,d \in \RR$ such that $bc- ad = 1$. We will see later that these indeed give solutions.

Suppose $p$ and $q$ satisfy the given equation; note that neither $p$ nor $q$ can be identically zero.
By subtracting the equations
\begin{align*}
p(x) q(x+1) - p(x+1) q(x) &= 1 \\
p(x-1) q(x) - p(x) q(x-1) &= 1,
\end{align*}
we obtain the equation
$$p(x) (q(x+1) + q(x-1)) = q(x) (p(x+1) + p(x-1)).$$
The original equation implies that $p(x)$ and $q(x)$ have no common nonconstant factor,
so $p(x)$ divides $p(x+1) + p(x-1)$. Since each of $p(x+1)$ and $p(x-1)$ has the same degree and leading
coefficient as $p$, we must have
$$p(x+1) + p(x-1) = 2p(x).$$
If we define the polynomials $r(x) = p(x+1) - p(x)$, $s(x) = q(x+1) - q(x)$,
we have $r(x+1) = r(x)$, and similarly $s(x+1) = s(x)$.
Put
$$a = r(0), b = p(0), c = s(0), d = q(0).$$
Then $r(x) = a, s(x) = c$ for all $x \in \ZZ$, and hence identically;
consequently, $p(x) = ax + b, q(x) = cx + d$ for all $x \in \ZZ$, and hence identically.
For $p$ and $q$ of this form,
$$p(x) q(x+1) - p(x+1) q(x) = bc - ad,$$
so we get a solution if and only if $bc-ad=1$, as claimed.