I'm close to certain it's just the product of the maximal tori of $O(m)$ and $O(n)$, but I can't quite prove it. I've tried the following:
$\newcommand{\mattwo}[4]{\left(\begin{array}{cc}#1\\#3\end{array}\right)}$
Any element of the Lorentz group can be decomposed into
$$\mattwo{P}{0}{0}{Q}\exp\mattwo{0}{X}{X^T}{0}$$
Where $P, Q$ are orthogonal block matrices, $P\in O(m)$, $Q\in O(n)$. So we can definitely pick maximal tori in $P$ and $Q$ (call them $T_1, T_2$), and $T_1\times T_2$ will definitely be a torus in $O(m,n)$.
To prove it's maximal, assume not. I try to find a contradiction using the fact that tori must be abelian. That's where things get a little ugly.
Take an arbitrary element $\mattwo{P'}{0}{0}{Q'}\exp\mattwo{0}{X}{X^T}{0}\in O(m,n)$ and supposed for the sake of contradiction that it lives in a larger torus containing $T_1\times T_2$. Then it must commute with every element in $T_1\times T_2$. This means for every $P\in T_1, Q\in T_2$, we must have
$$\mattwo{P}{0}{0}{Q}\mattwo{P'}{0}{0}{Q'}\exp\mattwo{0}{X}{X^T}{0}=\mattwo{P'}{0}{0}{Q'}\exp\mattwo{0}{X}{X^T}{0}\mattwo{P}{0}{0}{Q}$$
Knowing that $\exp\mattwo{0}{X}{X^T}{0}$ is symmetric, we can say $$\exp\mattwo{0}{X}{X^T}{0}=\mattwo{X_1}{X_2}{X_2^T}{X_3}$$
Which multiplies out to
$$\mattwo{PP'X_1}{PP'X_2}{QQ'X_2^T}{QQ'X_3}=\mattwo{P'X_1P}{P'X_2Q}{Q'X_2^TP}{Q'X_3Q}$$
and this gives us
\begin{align*} PP'X_1&=P'X_1P\\ PP'X_2&=P'X_2Q\\ QQ'X_2^T&=Q'X_2^TP\\ QQ'X_3&=Q'X_3Q\\ \end{align*}
which must hold for every $P\in T_1, Q\in T_2$.
This seems like it would be a strict enough set of four equations that would force $\exp\mattwo{0}{X}{X^T}{0}=\mattwo{1}{0}{0}{1}$, which would then force $P'\in T_1$, $Q'\in T_2$, but I can't for the life of me get the four equations to imply that.
It is known that $O(m)\times O(n)$ is a maximal compact subgroup of $O(m,n)$, and any maximal compact subgroup is conjugate to this one. A maximal torus, being compact, will therefore be contained in a conjugate of $O(m)\times O(n)$. Hence it will be conjugate to your $T_1\times T_2$.