Welcome to the site! Let $G = (V, E)$ be a undirected graph.
If $G$ is Hamiltonian, then there exists a simple cycle $C$ in $G$ containing every vertex, i.e. the length of $C$ is $|V|$.
Note that if you delete any edge $vw$ from $C$, you end up with a path of length $|V| - 1$ between $v$ and $w$.
Conversely, if you find a simple path $P$ of length $|V| - 1$ between two vertices $v, w$ connected by an edge, you can add the edge $vw$ to $P$ to obtain a Hamiltonian cycle (as $P$ cannot contain $vw$ due to being a simple path).
Also observe that the choice of $v$ and $w$ is arbitrary as long as they are connected by an edge.
So we can obtain the following reduction:
Starting from some graph $G = (V, E)$, pick some edge $vw \in E$ and delete it to obtain a new graph $G' = (V, E \setminus \{vw\})$.
Using the observations we made, it follows that $G'$ has a path $P$ between $v$ and $w$ of length $|V| - 1$ if and only if adding $vw$ to $P$ yields a Hamiltonian cycle in the original graph $G$.
As $G'$ can be computed from $G$ in polynomial time, we have a polynomial reduction $\mathrm{Hamiltonian Cycle} \leq_\mathsf{P} \mathrm{Long Path}$, thus showing that your problem is $\mathsf{NP}$-hard (because $\mathrm{Hamiltonian Cycle}$ is $\mathsf{NP}$-hard already).
To show that $\mathrm{Long Path}$ is in $\mathsf{NP}$ we note that a path in a graph can be efficiently encoded as a list of its edges and checking that a string actually encodes a path in the graph and then counting the number of edges to determine whether it is long enough can be done in polynomial time.
Hence we find that $\mathrm{Long Path}$ is $\mathsf{NP}$-complete.
