Example 6 If the polynomial $x^{2 k}+1+(x+1)^{2 k}$ is not divisible by $x^{2}+x+1$, find $k$.

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Let $w$ be a complex cube root of 1, then $w^{3}=1, w^{2}+w+1=0$, so $w$ is a root of $x^{2}+x+1=0$.
When $k=3 l$ ($l$ is a non-negative integer), $w^{2 k}+1+(w+1)^{2 k}=1+1+\left(-w^{2}\right)^{2 k}=3$;
When $k=3 l+1$, $w^{2 k}+1+(w+1)^{2 k}=w^{2}+1+w=0$;
When $k=3 l+2$, $w^{2 k}+1+(w+1)^{2 k}=w+1+w^{2}=0$.
That is, $w$ is not a root of $x^{2 k}+1+(x+1)^{2 k}=0$ if and only if $3 \mid k$, hence the answer to this problem is $3 \mid k$.