Leming and Jer found the same two solutions, with programming, but they gave no proof of existence of another one (or anothers).

My thought is : 

x^y - y^x = x+ y

x^y - x = y^x + y

x^y - x = y^x - y + y + y

x [x^(y-1) - 1] = y[y^(x-1) - 1] + 2y

The two expressions between parentheses are, respectively, divisible by (x-1) and by (y-1).

So, [x^(y-1) - 1] = k1.(x-1) for some integer value of k1.

And [y^(x-1) - 1] = k2.(y-1) for some integer value of k2.

Then :

x.k1(x-1) = y.k2.(y-1) + 2y

Or :

k1.x^2 - k1.x - [k2.y(y-1) + 2y) = 0

This is an second degree equation in x, so it has ONLY TWO roots, that are those already found.

So, I think that there are no more solutions to be searched. 

Am I wrong ?