First note that a reducible homogeneous polynomial necessarily splits into a product of homogeneous factors, so you can freely divide out by one of the variables. Thus your polynomial is irreducible if and only if $x^8 + y^8 + 1$ is irreducible.
Next you can use Eisenstein's criterion; think of $x^8 + y^8 + 1$ as a polynomial in $x$ with coefficients in $k[y]$. Since $\text{char}(k) \neq 2$ by assumption, the polynomial $y^8 + 1$ is squarefree, so it has a prime factor which divides it only once (exactly what this factor is depends on $k$), and by Eisenstein's criterion applied to this prime factor the conclusion follows.
For homogeneous polynomials in exactly three variables you can also use Bézout's theorem. If $k$ is algebraically closed and $f$ is reducible, then $f = 0$ defines at least two projective curves in $\mathbb{P}^2$, and by Bézout's theorem they intersect. At such an intersection point, all of the partial derivatives of $f$ will be equal to zero (exercise). So if you can show that there doesn't exist such a point, then $f$ must be irreducible. In this case, the partial derivatives are $8X_0^7, 8X_1^7, 8X_2^7$ which cannot simultaneously be zero if $\text{char}(k) \neq 2$.