The best-easiest solution is probably to use two MAX driver chips...
Another thing caught my eye though. In the MAX7219 datasheet, it says that the resistor between the ISET and V+ pins is for controlling the display brightness. Does this mean that with that single resistor between ISET and V+, I don't even need those 7 resistors on the SEG pins?
Right... But, because standard 7-segment displays have a common anode or common cathode, you can't wire the segmentsfrom different displays in series. And if you wire them in parallel without current limiting resistors, there is nothing to assure that the current is divided equally between them.
However, you might be able to fake-out the driver chip by using a resistor that "programs" very-high current, and then use a separate resistor on each segment to limit the actual current. But, that's something you'd have to experiment with, if you really think it's worth it... And, you might fry a driver chip with these "unauthorized" experiments.
Also, the datasheet says that a 10K resistor will limit the current on each SEG pin to about 40mA, what value resistor would be required to get that down to 10 or 20mA?
Page 10 of the data sheet:
Selecting RSET Resistor and
Using External Drivers
The current per segment is approximately 100 times
the current in ISET. To select RSET, see Table 11....
.