Solution:

Assume that the equation has a solution $(x_{0}, y_{0}, z_{0})$. Then $x_{0}^{2} + 2y_{0}^{2}$ is divisible by $7$. Since the remainders modulo $7$ of the perfect squares are $0, 1, 2$ and $4$, it follows that both $x_{0}$ and $y_{0}$ are divisible by $7$.

Then the left hand side of the given equation is divisible by $7^{2}$ and hence the number $\underbrace{11\ldots1}_{2005}$ is divisible by $7$. But this is a contradiction since $111111$ is divisible to $7$ and $2005 = 6 \cdot 334 + 1$.