Determine if the Diophantine equation $x^{2008}-y^{2008}=2^{2009}$ has any solutions.

The Diophantine equation to check is

$$x^{2008} - y^{2008} = 2^{2009} \tag{1}\label{eq1A}$$

It's clear the parity of $x$ and $y$ must be the same. Consider if they are both even, say $x = 2x'$ and $y = 2y'$. Then \eqref{eq1A} becomes, as user376343's question comment suggests,

$$2^{2008}(x')^{2008} - 2^{2008}(y')^{2008} = 2^{2009} \implies (x')^{2008} - (y')^{2008} = 2 \tag{2}\label{eq2A}$$

However, if $x' = \pm 1$ and $y' = 0$, then you get a result of $1$, while for any other values of $x'$ and $y'$ you will get, e.g., as suggested by the binomial theorem expansion, a difference of much more than $2008$ and, in particular, more than $2$.

This means that $x$ and $y$ must both be odd. Then, as you've shown, the left side of \eqref{eq1A} can be factored to get

$$(x^{1004} - y^{1004})(x^{1004} + y^{1004}) = 2^{2009} \tag{3}\label{eq3A}$$

Note $x^{1004} \equiv y^{1004} \equiv 1 \pmod{4} \implies x^{1004} + y^{1004} \equiv 2 \pmod{4}$. Thus, $x^{1004} + y^{1004}$ has just one factor of $2$. As such, unless $x, y = \pm 1$, which gives a value of $0$ in \eqref{eq1A}, then $x^{1004} + y^{1004}$ has an odd factor greater than $1$. However, the right side of \eqref{eq3A} is a power of $2$, so this is not possible.

In conclusion, there are no integer solutions to \eqref{eq1A}.