If $p|n$ odd prime divisor, $z=x^{an/p},t=y^{bn/p}$, then $(z+t)(z^{p-1}+...+t^{p-1}=2^{2010}.$ Therefore $t=2^k-z$ and $z^p+(2^k-z)^p=2^{2010}\to 2^k|z\to xy=0$ or $x^a=y^b=2^{k-1}$. It mean $n=2^k$, if $x^a\not =y^b$ (n had not odd prime divisors). Let $gcd(x^a,y^b)=2^m$. Then $mn<2010$, but $x^{an}+y^{bn}=2^{mn}\mod 2^{mn+3}$ if one of $\frac{x^a}{2^m},\frac{y^b}{2^m}$ is odd, and $x^{an}+y^{bn}=2^{mn+1}\mod 2^{mn+3}$ if both is odd. Therefore we had only solution $x^a=y^b=2^{k-1}\to (k-1)n=2009=7^2*41$. All solutions are $x=2^s,y=2^r$, were $s,r$ are divisors of $\frac{2009}{n}$. $n=2009,x=y=2,a=b=1$ $n=287,x^a=y^b=2^7$ - 4 solutions $n=49, x^a=y^b=2^{41}$ - 4 solutions $n=41, x^a=y^b=2^{49}$ - 9 solutions $n=7$,x^a=y^b=2^{287}$ - 36 solutions$. There are $54$ solutions.