Is there a solution to X^e + Y^e = Z^e where X,Y, and Z are all whole numbers?
X = 0, Y = Z works, but that’s trivial.
Just to be clear, you’re asking about e, the constant 2.718… ?
Yes, The base of the natural log.
More generally, is it even known if there any irrational exponents with solutions? (certain fractional exponents do work) What made you wonder about ‘e’ in particular?
One usually encounters e in the bottom of an exponentiation, not the top. Seeing something raised to the e power is is already a sign that things are going to get wonky.
Well, e is the magical constant 
when involved in logs and exponents.
Check out this graphical calculator:
Inputting y=x^1/x displays y max when x=e
X^n + Y^n = Z^n
n=1…simple arithmetic
n=2…Fermat’s Last Theorem (3:4:5, 5:12:13, etc)
n will not work for whole numbers greater than 2.
e is not a whole number…and neat mathematical stuff happens when you involve e.
Here is a pagethat indicates that no solution exists for rational exponents greater than 2, but that real values solutions definitely exist. That said I would be very surprised if there existed solutions for e.
I am more than impressed that you found that. Very nice.
This does not answer the OP (I very much doubt that any knows the answer to that one), but here is one thing to think about. Since 6^3 + 7^3 > 8^3 and 6^4 + 7^4 < 8^4, the function f(x) = 6^x + 7^x - 8^x is positive when x = 3 and negative when x = 4 and therefore 0 at some point in the interval 3 < x < 4. Almost surely irrational and very likely transcendental.
BTW, the equation x^2 + y^2 = z^2 is not Fermat’s equation. It was known to Euclid to have infinitely many solutions. He knew that if you chose numbers m > n with no common divisor and one even, then x = 2mn, y = m^2 - n^2 and z = m^2 + n^2 you always got a “primitive” solution (one in which x, y and z had no common divisor). It is unclear if he realized that this gives all the primitive solutions (but it does). The Fermat equation is x^n + y^n = z^n for n > 2.