Yes you can,
num = 2^74207281 - 1;
Do[If[Mod[num, n] == 0, Print["Fail"]; Abort[]], {n, 2,
Ceiling[Sqrt[num]], 2}];
leave that running for a while ;-) If you see a Fail, then it is not a prime...
In all seriousness, there are better (faster) ways of checking it. The above code takes forever. Have a look at this video:
https://www.youtube.com/watch?v=lEvXcTYqtKU
they explain a more reasonable way of checking... but still this takes a LONG time!!
