Let $x \le y$. $x=1$ gives us an obvious solution $y=2007$.

In what follows, suppose that $2 \le x \le y$. We can check for small values of $x$:

$2^{10} = 1024 < 2008 < 2^{11} = 2048$, $3^6 = 729 < 2008 < 3^7 = 2187$,
$4^5 = 1024 < 2008 < 4^6 = 4096$, $5^4 = 625 < 2008 < 5^5 = 3125$, $6^4 = 1296 < 2008 < 6^5 = 7776$, $7^3 = 343 < 2008 < 7^4 = 2401$, $8^3 = 512 < 2008 < 8^4 = 4096$, $9^3 = 729 < 2008 < 9^4 = 6561$, $10^3 = 1000 < 2008 < 10^4 = 10000$.

It is clear, that there is no need to continue. That is, it follows from the relations $2^{10} + 10^2 = 1124 < 2008$, $3^6 + 6^3 = 945 < 2008$, $4^5 + 5^4 = 1649 < 2008$ that there are no more new solutions (for further values of $x$, the condition $2 \le x \le y$ breaks).