Longest Hamiltonian Cycle probably isn't known to be in NP. We can certainly check in deterministic polynomial time if proposed solution is a Hamiltonian Cycle but, as you say, it doesn't seem possible to check that it's the longest one.
We usually consider the decision version of this kind of maximization problem to be, "Given an instance $X$ and a target $t$, is there a solution with value at least $t$?" So, for Longest Hamiltonian Cycle, we're given a graph $G$ and a number $t$ and we're asked if there's a Hamiltonian cycle of length at least $t$.
Clearly, Hamiltonian Cycle for unweighted graphs is reducible to this: set all the weights to $1$ and ask if there's a Hamiltonian cycle of length at least $1$ (or $n$, if you prefer).
Longest Hamiltonian Cycle is also reducible to it, though using Turing reductions rather than the many-one reductions we use for NP-completeness. We can decide LHC by binary search. Assuming all the edge weights are non-negative, we know that the longest Hamiltonian cycle must have length at least zero, and that it can't be longer than $M$, the sum of the longest edge out of each vertex. We can do binary search on the interval $[0,M]$ with $\log M$ calls to a subroutine that decides the "LHC at least $t$?" problem, and $M$ can't be more than exponential in the length of the input, so this reduction runs in polynomial time.