Answer. $2^{14} - 2^7 = 16256$.

First solution. If all sequences whose number of letters is at least $7$ and at most $13$ are words, then, obviously, the condition of the problem is satisfied; in this case, the number of such words is $2^7 + \dots + 2^{13} = 2^{14} - 2^7$. It remains to show that this number is the largest possible.

The total number of sequences of length not exceeding $13$ is $2 + 2^2 + \dots + 2^{13} = 2^{14} - 2$. If there are no $7$-letter words among the words in the language, then the total number of words does not exceed $2^{14} - 2 - 2^7 < 2^{14} - 2^7$. Suppose, on the contrary, that the language contains a $7$-letter word $s$. Then for each word $t$ consisting of $6$ or fewer letters, the sequence of letters $st$ cannot be a word, and all sequences of the form $st$ are obviously distinct. Therefore, if the language has $k$ words of $6$ or fewer letters, then the number of words of at least $7$ letters does not exceed $(2^7 + \dots + 2^{13}) - k = 2^{14} - 2^7 - k$. Thus, the total number of words does not exceed $k + (2^{14} - 2^7 - k) = 2^{14} - 2^7$, which is what was required to prove.

Second solution. Here is another proof that the number of words in the language does not exceed $2^{14} - 2^7$. Let $A$ be the set of all sequences of $6$ or fewer letters, and $B$ be the set of all $7$-letter sequences. Then $A$ contains $2 + 2^2 + \dots + 2^6 = 2^7 - 2$ sequences, and $B$ contains $2^7 > 2^7 - 2$ sequences. Therefore, it is possible to assign to each sequence $a \in A$ a sequence $b_a \in B$ so that all sequences $b_a$ are distinct. Note that then all sequences of the form $ab_a$ are also distinct (since their $7$-letter endings are distinct).

By the condition, in each of the $2^7 - 2$ triples $(a, b_a, ab_a)$ there are at most two words of the language. Therefore, at least $2^7 - 2$ sequences of $13$ or fewer letters are not words, and the total number of words does not exceed $(2^{14} - 2) - (2^7 - 2) = 2^{14} - 2^7$.