$\bullet\ \textbf{Question}$
One can find equivalent products of consecutive integers such as $$8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14=63\cdot64\cdot65\cdot66.$$ Other solutions of this have been posted already (Equal products of consecutive integers). But the implied meaning of 'trivial' solutions in other posts seems unsatisfactory -- I think I can convince you. There appear to be exactly $4$ non-trivial solutions after adjusting our vocabulary. One of which is the previous equation. I'm curious; are their any other non-trivial solutions?
$\bullet\ \textbf{Triviality}$
Firstly, let's use the notation $a^{(b)}=\prod_{k=0}^{b-1}(a+k)$ to denote rising factorial. The previous equation then becomes $8^{(7)}=63^{(4)}$.
All solutions come in pairs -- one overlapping and one disjoint. For example, if we take the solution $$2^{(5)}=2\cdot3\cdot4\cdot5\cdot6=8\cdot9\cdot10=8^{(3)},$$ which we will call disjoint because no integer appears on both sides, we can create an overlapping solution multiplying both sides by $7$: $$2^{(6)}=2\cdot3\cdot4\cdot5\cdot6\cdot\textbf{7}=\textbf{7}\cdot8\cdot9\cdot10=7^{(4)}.$$ This duality means that boring solutions like $2^{(2)}=2\cdot3=6^{(1)}$ can be disguised as interesting by transforming them from disjoint to overlapping: $$2^{(2)}=6^{(1)}\quad\rightarrow\quad 2^{(4)}=4^{(3)}.$$ For example, two answers on a previous post gave $3^{(9)}=5^{(8)}$ as an "interesting" solution. But this is simply the overlapping dual of $3\cdot 4 = 12$.
In general, if $a^{(b)}=c^{(d)}$ is a solution, then so is its dual: $$a^{(c-a)}=(a+b)^{(c+d-a-b)}.$$
So here we define a trivial solution either 1) having overlap or 2) having only one term on one side (previous inquiries seemed to only have considered #2).
$\bullet\ \textbf{Non-trivial Solutions}$
There are $4$ known non-trivial solutions under this definition: $$2^{(5)}=8^{(3)},\quad 5^{(3)}=14^{(2)},\quad 19^{(4)}=55^{(3)},\quad\text{and}\quad 8^{(7)}=63^{(4)}.$$ I ran a computer search on equivalent $a^{(b)}$ for $a<100,000$ and $b<50$ and found nothing new (https://repl.it/@onnomc/EqualProductsOfConsecutiveIntegers).
$\bullet\ \textbf{Proof Attempts}$
One attempt is to note $a^{(b)}=\frac{(a+b-1)!}{(a-1)!}$ and that therefore solutions to $a^{(b)}=c^{(d)}$ are also solutions to $A!B!=C!D!$. But apparently even $A!B!=C!$ has not yet been solved (On the factorial equations $A! B! =C!$ and $A!B!C! = D!$).
I thought also to use Stirling's approximation since $$(a+1)^{(b)}=\frac{(a+b)!}{a!}\approx \Big(\frac{a+b}{a}\Big)^{a+1/2}\Big(\frac{a+b}{e}\Big)^b$$ and ... well, hope for the best. I made no progress here.
I tried also bounding the number of distinct primes dividing $a^{(b)}$. We can bound $c$ from below since $$a^b<a^{(b)}=c^{(d)}<(c+d-1)^d\quad\Rightarrow\quad c>a^{b/d}-d+1.$$ And of course, the non-overlapping criteria gives $c\ge a+b$. Thus if the prime bound rises fast enough, we can show that we are trying to fit too many prime divisors into $c(c+1)...(c+d-1)$. The furthest I got down this trail was finding $r$ consecutive integers with a product divisible by at most $r$ primes. For example, the largest solution for $r=5$ is $c=24$ since $$24\cdot25\cdot26\cdot27\cdot28=2^6\cdot3^4\cdot5^2\cdot7\cdot13.$$ The largest solution for $r=42$ seems to be $c=2175$. But I cannot prove these are the largest such integers -- only that they appear so empirically (https://repl.it/@onnomc/ConsecutiveIntegersOfFewPrimes). Interestingly, the case $r=2$ yields $c=+\infty$ if there are infinitely many Mersenne primes.