The Problems of Ukrainian Authors

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).