If I had the following equation, it is easy to convert to only NAND and NOT gates
$$(A\land \neg B) \lor (C\land B) \iff (A\text{ NAND } \neg B) \text{ NAND } (C\text{ NAND } B)$$
Just replace all the ors and ands with NANDS.
I understand the logic behind the NAND, before I used to do it using laws when I was learning, but now I just replace it all with NAND.
Similarly, is there a trick for converting to NOR?
Yes. You discovered that if you have a statement that is a disjunction of two disjuncts, and where each of the disjuncts is a conjunction of two literals (as in your example), then you can replace all $\land$'s and $\lor$'s with NAND's.
So, that tells me that if you have a statement that is a conjunction of two conjunct, and where each of the conjuncts is a isjunction of two literals, then you can replace all $\lor$'s and $\land$'s with NOR's.
How do I know this? Because of the duality theorem!
Duality Theorem (stated informally)
Whenever you have some equivalence principle involving a bunch of binary operators, then you can find a 'dual' equivalence principle by replacing all binary operators with their duals.
The Duality Theorem is why Commutation, Association, DeMorgan, Distribution, Absorption, Idempotence, Reduction, etc. etc. all come in pairs
Applied to your case: Given as the $\land $ and $\lor$ are each other's dual, and given that NAND and NOR are each other's dual, you get exactly what I wrote at the beginning of the post.
OF course, that this dual principle is true for the NOR's is easily verified in and of itself.