Advance Happy New Year 2015!!

The fourth power of x leaves a residue of 0 or 1 mod 4. However, 2015 is congruent to 3 mod 4. Thus, no values of x and y satisfy the equation.

$x^4\equiv 0,1 \pmod{4}$

$y^4\equiv 0,1 \pmod{4}$

$\Rightarrow$ $(x^4+y^4)\equiv 0,1,2 \pmod{4}$

$\Rightarrow$ But $2015\equiv 3 \pmod{4}$

$\Rightarrow$ So no such x or y exists. Therefore the ans is $\boxed{0}$

Key technique:What we are going to do here is divide 'x' and 'y' into various categories based on the remainder they leave when divided by 4.Keep in mind that 2015 leaves a remainder of 3 when divided by4. $\begin{cases}\underline{x}\\4*a\\4*a+1\\4*a+2\\4*a+3\end{cases}\begin{cases}\underline{b}\\4*b\\4*b+1\\4*b+2\\4*b+3\end{cases}$ Now, $x^{4}$ and $y^{4}$ can leave remainders, $\begin{cases}\\0(4*b,4*a)\\1(4*b+1,4*a+1)\\0(4*b+2,4*a+2)\\1(4*b+3,4*b+3)\end{cases}.$ Thus, $L.H.S$ can leave remainders, $0,1,2$ when divide by $4$ but the $R.H.S$ leaves a remainder of $3$ when divided by $4.$ Thus,there are no possible pairs.

$\sqrt[4]{2015} =6.6999. ~~\implies~~ x,y<7.\\However~ 2015~is ~odd~ \therefore ~one ~of~x^4~and~y^4~ must~be~odd~other~even.\\Starting~with\\~x=1,~ 2015-1^4 =2014~is ~not~a~fourth~power~of~an ~integer.\\~~~x=3,~ 2015-3^4=1934~is ~not~a~fourth~power~of~an ~integer.\\~~~~~x=5,~2015-5^4=390~is ~not~a~fourth~power~of~an ~integer. \\x<7 so~ there~ are~no~othere~integers~to~check. \\\therefore~no~integer~x,y~satisfy~the~equation.$

Taking mod $4$ the L.H.S should be 1 or 0 or 2,but the right side is 3 modulo 4.contradiction,such number don't exist.

Hence,the answer is $\boxed{0}$ .

the exponent could even be 2,the exact same argument would still work.