Lets start with n=2.  Then the left side factors into (x-y)*(x+y).  Clearly, x-y will always be smaller than x+y.  2015 has four positive factorizations: 1*2015, 5*403, 13*155, 31*65.

Then there are four cases:

x-y=1 and x+y=2015 yield x=1013, y=1012.

x-y=5 and x+y=403 yield x=204, y=199.

x-y=13 and x+y=155 yield x=84, y=71.

x-y=65 and x+y=65 yield x=48, y=17.

Onto the n=3 case.  Then the left side factors into (x-y)*(x^2+xy+y^2).  In this case we will note (x-y)^2 < x^2+xy+y^2.  That means we only want factorizations of 2105 where the smaller number is less than the cube root of 2015, x-y<cbrt(2015)=12.6.  There are only two such factorizations 1*2015 and 5*403.

x-y=1 and x^2+xy+y^2=2015 implies (y+1)^2 + (y+1)*y + y^2 = 2015, which has roots of y=(-3+/-sqrt(24177))/6. These are irrational and must be discarded.

x-y=5 and x^2+xy+y^2=403 implies (y+5)^2 + (y+5)*y + y^2 = 2015, which has roots of -14 and 9.  -14 needs to be discarded which leaves y=9 and x=14 as a solution.

Bigger cases!  First I will note that 2^11-1 = 2047 is already bigger than 2015, so that limits the exponent to be at most 10.

Next thing is to see that the largest x could be is when y is one less than x.  We'll use this fact to get us an upper bound on x in terms of n.

So then write x as (x-1/2)+1/2 and y=x-1 as (x-1/2)-1/2.  Now apply the binomial expansion:

[(x-1/2)+1/2]^2 = (x-1/2)^2 + nC1*(x-1/2)^(n-1)*(1/2) + nC2*(x-1/2)^(n-2)*(1/2)^2 + nC3*(x-1/2)^(n-3)*(1/2)^3 + ....

[(x-1/2)+1/2]^2 = (x-1/2)^2 - nC1*(x-1/2)^(n-1)*(1/2) + nC2*(x-1/2)^(n-2)*(1/2)^2 - nC3*(x-1/2)^(n-3)*(1/2)^3 + ....

Now subtract these expressions to get

2*nC1*(x-1/2)^(n-1)*(1/2) + 2*nC3*(x-1/2)^(n-3)*(1/2)^3 + ... = 2015.

We can make an inequality by dropping all but the first term (and simplify a bit): n*(x-1/2)^(n-1) < 2015 implies x < (2015/n)^(1/(n-1))+1/2

A lower bound for x is easier, just drop y^n: x^n>2015 implies x>2015^(1/n).

I will round the lower estimate down and the upper estimate up to not accidentally lose a boundary case.

For n=4 we have x=6,7,8,9

For n=5 we have x=4,5

For n=6 we have x=3,4

For n=7 we have x=2,3,4

For n=8, 9 or 10 we have x=2,3

x-y is still a factor, but with x being so small the only factors that make sense are x-y=1 for all cases or x-y=5 for n=4 case. In all there are 21 specific cases to check: (n,x,y)=(4,6,5), (4,6,1), (4,7,6), (4,7,2), (4,8,7), (4,8,3), (4,9,9), (4,9,4), (5,4,3), (5,5,4), (6,3,2), (6,4,3), (7,2,1), (7,3,2), (7,4,3), (8,2,1), (8,3,2), (9,2,1), (9,3,2), (10,2,1), (10,3,2).

This is a small enough set that a spreadsheet formula can calculate the values of x^n-y^n.  But none of these evaluate to 2015.

So, after all this we have five triplets (x,y,n) = {(1008,1007,2),  (204,199,2), (84,71,2), (48,17,2), (14,9,3)}.