We cannot have $B = 0$ or $D = 0$, as this would imply $H = 0$. Similarly, we cannot have $B = 1$ or $D = 1$, as this would imply $H = D$ or $H = B$. For a solution $EFGH$ to be bigger than $6840$ one of the numbers needs to be greater than $86$ because otherwise the product would not exceed $79 \cdot 86 = 6794$. If we assume $AB < CD$ then $CD$ needs to be one of the following numbers $87, 89, 92, 93, 94, 95, 96, 97, 98$.

For each of these we can find the possible values of $AB$ which would make the product greater than $6840$, i.e. we work with the inequality
$$CD > AB \geq \left\lfloor \frac{6840}{CD} \right\rfloor .$$
For $C = 8$ we need to have $AB < 80$. If $CD = 87$ we even need to have $AB < 70$ to avoid repeated digits. But $70 \cdot 87 < 70 \cdot 90 = 6300 < 6840$. If $CD = 89$, we get $\lfloor \frac{6840}{CD} \rfloor = 77$ and none of $77, 78, 79$ is possible for $AB$ because of repeated digits.

Note that when $AB < 70$ we have $AB \times CD < 70 \cdot 98 < 7000$ and $E = 6$ whenever $AB \times CD > 6840$. Hence, we need to have $AB \ge 72$ to avoid repeated digits. Suppose now that $CD \ge 92$, then $A = 7$ or $A = 8$.

If $A = 7$, we have $AB \times CD < 80 \cdot 98 < 8000$ which implies that $E = 6$ whenever we have a solution with $AB \times CD > 6840$. This implies
$$72 \leq AB \leq \left\lfloor \frac{7000}{CD} \right\rfloor.$$
We now list all possibilities in a table where we have excluded some by keeping in mind $E = 6$ and by checking the last digits.

For example, when $CD = 92$ we excluded $AB = 73$ because the last digit of the product would be $6$. We excluded $CD = 96$ and $CD = 97$ because $E = 6$ and $A = 7$. We now check that none of these possibilities leads to a solution. Indeed,
$75 \cdot 92 = 6900$
$74 \cdot 92 = 6808$
$74 \cdot 93 = 6882$
$73 \cdot 94 = 6862$
$72 \cdot 94 = 6768$
do not give solutions, except of course for $72 \cdot 95 = 6840$.

If $A = 8$, we have $AB \times CD < 90 \cdot 98 < 9000$ and $AB \times CD \geq 82 \cdot 92 = 7544$ which implies that $E$ must be $7$. As $F$ can now be at most $6$, we actually have $AB \times CD < 7700$, hence
$$82 \leq AB \leq \left\lfloor \frac{7700}{CD} \right\rfloor.$$
Because $94 \cdot 82 = 7708$, the only values left for $CD$ are $92$ and $93$. If $CD = 93$, $7700/93 < 83$ and the only possibility is $AB = 82$. But $82 \cdot 93 = 7626$, no solution. If $CD = 92$, $AB \neq 82$ and $7700/92 < 84$, hence $AB$ could only be $83$. But $83 \cdot 92 = 7636$. Thus $6840$ is the largest possible $EFGH$.