(I'm assuming that the 2 k's are different; for simplicity purposes, I'm gonna call them 3a+1 and 6b+1)
We can prove this statement by contradiction.
Notice that if a is even, we can represent it as 2(3b)+1=6b+1 for some integer b. Let's say that a is not even.
If a is odd, 3a will also be odd, but if you add one to an odd number, it automatically becomes even, meaning that 3a+1 will always be even for odd values of a. If a number is even, it will always be composite because it has a factor of 2 (except if it is 2 itself, but it's irrelevant since it's unobtainable). Therefore, a must be even, meaning that it must be expressible as 6b+1 for some integer b.