Let $m,n,d$ all be positive integers with $m+n>2.$
Proposition: There exist prime numbers, $p_1,p_2,\ldots,p_m;\ \ q_1, q_2,\ldots,q_n,$ all distinct, such that $\displaystyle\prod_{i=1}^{m} p_i = \prod_{i=1}^{n} q_i + d. $
Other than obvious observations such as, if $d$ is odd then one of the prime factors must be $2,$ I do not have any great idea for how to approach such a problem. I haven't checked properly, but I doubt the prime number theorem will be useful.
This is an incomplete answer, relating your question to Dickson's conjecture which is presently unproven. It does suggest that your claim holds when $m,n>1$, and I'm sure that with some effort or a clever argument this can be turned into a proof. Unfortunately I'm in a hurry.
Let $m$, $n$ and $d$ be given. Without loss of generality I'll assume that $p_i<p_j$ and $q_i<q_j$ whenever $i<j$.
Pick arbitrary distinct primes $p_1,\ldots,p_{m-1}$ and $q_1,\ldots,q_{n-1}$ that are coprime to $d$, and set $$P:=\prod_{i=1}^{m-1}p_i\qquad\text{ and }\qquad Q:=\prod_{i=1}^{n-1}q_i.$$ Then we want to solve $$Pp_m-Qq_n=d,$$ where $p_m$ and $q_n$ are primes with $p_m>p_{m-1}$ and $q_n>q_{n-1}$. Because $P$ and $Q$ are coprime there exist positive integers $x$ and $y$ such that $$Px-Qy=d,$$ and then of course $(x+kQ,y+kP)$ is another solution for any integer $k$. Because $P$ and $Q$ are coprime to $d$, also $x$ is coprime to $Q$ and $y$ is coprime to $P$. By Dirichlet's theorem the two arithmetic sequences $x+kQ$ and $y+kP$ both contain infinitely many primes. Furthermore the asymptotic densities of primes in these sequences are $\varphi(Q)^{-1}$ and $\varphi(P)^{-1}$, respectively, where $\varphi$ denotes the totient function.
Heuristically one would expect that the integers $k$ for which $x+kP$ is prime are 'independent' of the integers $k'$ for which $y+k'Q$ is prime, and hence that there are in fact infinitely many integers $k$ for which $x+kP$ and $y+kQ$ are both prime. This is also the gist of Dickson's conjecture, which is still unproven.
When $m,n>1$ we are still free to choose $P$ and $Q$, to some extent, giving a lot of room to pin down a solution.
