By checking the parity of the prime factorization of all the integers in $T$, we partition them into disjoint triples, whose products are all squares:
$$\{8, 13, 26\}, \{11, 18, 22\}, \{14, 21, 24\}, \{12, 15, 20\}, \{9, 16, 25\}.$$
If $|T| \ge 16$, at least one of these triples will be in $T$, a contradiction.

We can easily check that the subset (obtained by strategically removing one element $8$, $18$, $24$, $12$, $9$ from each of the triples above):
$$\{7, 10, 11, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 26\}$$
satisfies the condition given. Thus the maximum number of elements in $T$ is $15$.