If $6^{3k} + 1$ is factorable, then it would be easy to show that it isn't prime. I know I could also represent the formula as $6^{3k} + 1^{3k}$, but the fact that the exponent could be even or odd is giving me some difficulty.
It's easy enough when the exponent is odd as I can treat it as $6^{3k} - ({-1})^{3k}$ and use the difference of powers to factor, but this doesn't apply when the exponent is even. Is there some way I can use sum of cubes to help with it?
You can factorise it as suggested by the comments (which is probably the standard way to answer the question), but you had a very good idea.
Assuming $3k$ is odd, as you noted, we can write:
$$6^{3k} - (-1)^{3k} = (6+1)(6^{3k-1}- 6^{3k-2} + \cdots + 1)$$
Note that, in particular, $7 | (6^{3k} + 1)$ in this case. To derive this another way, note that if $3k$ is odd:
$$6^{3k} + 1 \equiv (-1)^{3k} + 1 \equiv 0 \text{ mod } 7$$
So what do we do if $3k$ is even? Well, note that $2 \nmid 3$, so by Bézout, $2|k$. Hence, we can consider:
$$6^{3k} + 1 = (36)^{3 \cdot \frac{k}{2}} + 1$$
If $3 \cdot \frac{k}{2}$ is odd, then we can apply exactly the same argument. We find that:
$$6^{3k} + 1 = (36)^{3 \cdot \frac{k}{2}} + 1 \equiv (-1)^{3 \cdot \frac{k}{2}}+1 \equiv 0 \text{ mod } 37$$
where you may note that we have chosen $37 = 6^2 + 1$.
By induction, we deduce that if $k = 2^{s} m $ such that $m$ is odd, then:
$$(6^{s+1} + 1) \text{ divides } (6^{3k} + 1)$$
In particular, $6^{3k} + 1$ is not prime.
Remark: note that we used nothing special about $6$, and we only used that $3$ is odd. Thus, the result can be generalised much more broadly.