$m > 2, n > 2$ $$T_{m,n} = <V,E> $$ $$V = \{0,1,...,m-1\}\times\{0,1,n-1\} $$ $$E = \{\{<i,j>, <(i + 1)\pmod m, j>\} : <i,j>\in V\} \\ \cup \{\{<i,j>, <i, (j+1)\pmod n>\} : <i,j>\in V \} $$
For example:
Every degree is $4$. So according to Ore's theorem let's look at two not linked nodes: sum of their degrees is equal to $8$. So graph is planar if and only if $|V| = m\cdot n \le 8$ Thus, answer is $$(m,n) \in \emptyset $$
Is it properly solution ?
Ore's Theorem says let $G$ be a graph of order $p$ and if, for every pair of nonadjacent vertices $v$ and $w$ we have $\deg{v}+\deg{w}\ge p$, then $G$ is Hamiltonian. You cannot apply this theorem when your 4-regular graph has $p=m\cdot n> 8$, but that does not mean that it isn't Hamiltonian.
Looking at the graph you have defined, it appears it is always Hamiltonian. The Hamiltonian cycle depends upon the parity of the number of vertical vertices, but the cycle always exists. Try and find it!