For $K_{4,4}$ you can refer to this question.
For $K_{2,3}$, Suppose that you have an Hamiltonian cycle $(v_1v_2v_3v_4v_5)$.
Your graph is composed of two independent set of vertices $A$ and $B$, with $\vert A\vert=3$ and $\vert B\vert = 2$. Without loss of generality, we can suppose that $v_1\in A$ (otherwise look at the the cycle $(v_2v_3v_4v_5v_1)$).
Then
$$v_1\in A \Rightarrow v_2\in B \Rightarrow v_3\in A \Rightarrow v_4\in B \Rightarrow v_5\in A \Rightarrow v_1\in B $$
Hence a contradiction. In general, for any $n\neq m$, $K_{n,m}$ is not Hamiltonian (but Dirac's theorem is not a proof)
The paper "Randomly Traceable Graphs" by Gary Chartrand and Hudson V. Kronk tackles this question, and confirms that the three types of graph ($C_n$, $K_n$, $K_{m,m}$) are the only ones:
https://epubs.siam.org/doi/abs/10.1137/0116056
Randomly Traceble graphs are graphs where every random self-avoiding path eventually visits all vertices (i.e. becomes a Hamiltonian path).
Randomly Hamiltonian graphs are Randomly Traceble graphs where every one of those random Hamiltonian paths can be closed to become a Hamiltonian cycle.
The paper first proves that all Randomly Traceble graphs with $n\ge3$ are randomly Hamiltonian. This is simple: Given any randomly traceable graph with a Hamiltonian path $v_1$, $v_2$, ..., $v_n$. The path $v_2$, ..., $v_n$, must be extendable to a Hamiltonian path since the graph is Randomly Traceable. Therefore $v_n$ and $v_1$ are connected by an edge, and any Hamiltonian path can be closed to become a Hamiltonian cycle. In the rest of the paper this trick of proving an edge must be present by constructing a hamiltonian path between its two vertices is repeatedly used.
The paper then views any Randomly Traceable graph as an outer (Hamiltonian) n-cycle possibly with diagonals.
If it has no diagonals, then it is just a cycle graph $C_n$.
If it has triangles (formed by two outer edges and one diagonal) then it turns out that all diagonals must be present and it is a complete graph $K_n$.
If instead it has 4-cycles (formed by three outer edges and one diagonal) then it turns out that it must be $K_{n/2,n/2}$ with $n$ even.
If it has larger cycles, then it will also have triangles, so no other types of randomly Hamiltonian graphs exist.
Best Answer
Let's look at the definition of a Hamiltonian cycle:
It is a cycle (i.e. closed path. You see, no repetition of vertices except for the starting point, no repetition of edges) that goes through all the vertices.
So if we fix $n$ to be the number of vertices, each Hamiltonian cycle has the length $n$.
Now suppose there is a cycle of length more than $n$. Then it has to visit some vertex multiple times, but this is in contradiction with the definition of a cycle.