Solution:

For $x < 0$ or $z \leq 0$ the only solution is the trivial one $(0, 0, 0)$. From now on let $z > 0$. Since $2008 = 2^{3} \cdot 251$, both sides of the equation are divisible by $251$. Suppose that $x = 2x_{1}$ is even. Then $\left(2^{x_{1}}\right)^{2} \equiv -\left(y^{2}\right)^{2} \pmod{251}$, which upon raising to the $125$-th power gives $1 \equiv \left(2^{x_{1}}\right)^{250} \equiv -\left(y^{2}\right)^{250} \equiv -1$ by Fermat's little theorem, which is impossible. Therefore, $x$ must be odd.

It is clear that $y$ is even. Let us write $y = 2^{u} y_{1}$ for odd $y_{1}$. We have
$$2^{2x} 3^{x} + 2^{4u} y_{1}^{4} = 2^{3z} 251^{z}$$
Since $2x \neq 4u$ because $x$ is odd, the highest power of two dividing the left-hand side is $2^{2x}$ or $2^{4u}$, while the highest power of two dividing the right-hand side equals $2^{3z}$, whence $3z = 2x$ or $3z = 4u$. We shall show that in neither of these two cases does the given equation have solutions.

i. $3z = 2x < 4u$; hence, $2 \mid z$. Dividing by $2^{2x}$ gives $3^{x} + 2^{4u-2x} y_{1}^{4} = 251^{z}$ which is impossible because the left-hand side is of the form $4k+3$ (since $2 \nmid x$), while the right-hand side is of the form $4k+1$ ($k \in \mathbb{N}$).

ii. $3z = 4u < 2x$; again $2 \mid z$. Dividing by $2^{4u}$ gives $2^{2x-4u} 3^{x} + y_{1}^{4} = 251^{z}$. The right-hand side is of the form $5k+1$, so for $5 \nmid y_{1}$ we have $y_{1}^{4} \equiv 1$ and $2^{2x-4u} 3^{x} \equiv 0 \pmod{5}$ which is impossible, while for $5 \mid y_{1}$ we have $1 \equiv 2^{2x-4u} 3^{x} \equiv \pm 3^{x} \equiv \pm 3 \pmod{5}$ since $2 \nmid x$, again impossible.

Solution:

For even $x$ the left-hand side of the equation is of the form $a^{2} + b^{2}$, and for odd $x$ the left-hand side is of the form $a^{2} + 3b^{2}$. However, since $-1$ and $-3$ are quadratic non-residues modulo $251$, neither $a^{2} + b^{2}$ nor $a^{2} + 3b^{2}$ can be divisible by $251$ if $251 \nmid a, b$. Therefore, the given equation has no integer solutions for $z \geq 0$.