Let $\mu$ denote the expectation of the expected number of rolls that are still needed if $2$ rolls have passed with distinct result.
Let $\nu$ denote the expectation of the expected number of rolls that are still needed if $2$ rolls have passed with equal result.
Then the expectation is: $$2+\frac56\mu+\frac16\nu$$
Here $\frac56$ is the probability that the first two numbers are distinct and $\frac16$ is the probability that they are equal.
Secondly we have the relations:
$$\mu=\frac56(1+\mu)+\frac16(1+\nu)=1+\frac56\mu+\frac16\nu\tag1$$ and: $$\nu=\frac161+\frac56(1+\mu)=1+\frac56\mu\tag2$$
The relations $(1)$ and $(2)$ lead easily to: $\mu=42$ and $\nu=36$.
Then $$2+\frac56\mu+\frac16\nu=43$$ is the final answer.
addendum:
(in the answers uptil now it was not used that you allready learned something).
Let $Y$ denote the number of rolls required to see three dice of the same number in succession and let $X$ denote the number of rolls required to see three dice with number $6$ in succession.
Then: $$Y\text{ and }\frac16X+\frac56(X+Y)=X+\frac56Y\text{ must have equal distribution.}\tag3$$
Here $\frac16$ is the probability of the event that the first time that three dice give the same number in succession they show number $6$ and $\frac56$ is the probability that do not show a number $6$.
$(3)$ rests on the observation that - if for the first time three equal numbers show up in succession - we are ready if $6$ happens to be that number and must actually start over again (with $X$ throws in our pocket) if not.
So we find $\mathbb EY=\mathbb EX+\frac56\mathbb EY$ or equivalently:$$\mathbb EX=\frac16\mathbb EY$$
You allready learned that $\mathbb EY=258$ and making use of that knowledge you find $$\mathbb EX=258/6=43$$ This confirms your thinking.