Coincidences within results of each player are possible.
Example of "successful" event (n=m=3): {1,1,6} and {3,4,5} (no coincidences)
Example of "unsuccessful" event (n=m=3): {1,2,6} and {2,4,5} (1 coincidence - "2")
My guess is that $$prob=(5/6)^{m*n}$$, but when I run a simulation program, I get a result slightly different from "theoretical" one.
Your expected result has two problems: There are $nm$ pairs of dice, each of which might match, so you might expect the chance to be $(\frac 56)^{nm}$. The equivalent for $s$ sided dice of $(\frac{s-1}s)^{mn}$ would be pretty close if $s$ is very large. The other is that when one player has matching dice, there are less opportunities for a match.
What you need to do is calculate the chance that each player has $1,2,...,6$ different numbers. For $n=m=3$, each player has one number $\frac 6{216}$ of the time, two different numbers $\frac {90}{216}$ of the time, and three different numbers $\frac {120}{216}$ of the time. The chance of no match is then $\frac 1{216^2}(6*6*\frac 56+2*6*90*\frac 46+2*6*120*\frac 36+90*90*\frac{6}{15}+2*90*120*\frac 3{15}+120*120*\frac 1{20})=\frac{1625}{7776}\approx 0.2090$