First, observe that the square of no prime is triangular or a Fibonacci number. Therefore, $n$ must be squarefree or the square of a prime $p$. But in the latter case, there are only $2$ divisors, contradicting $\tau(n) = |F|+|T| +1 \geqslant 4$. Suppose $n = p_1p_2\dots p_k$. We consider two cases: 1) $1 \in T$ In this case, for some positive integer $u \geqslant 2$ we know that all triangular numbers $1, 3, \dots, \frac{u(u+1)}{2}$ divide $n$. If $u \geq 3$, then $6 \mid n$ and it follows that $2 \in F$. Since $1 \not \in F$ it follows that $3 \in F$ ($|F| \geqslant 2$) but then $3$ is in both $F, T$ a contradiction. Thus $u = 2$ and we must have only $1, 3$ in $T$ and the other divisors must be in $F$. Observe that $n$ must be odd now. As a result, $F$ cannot contain any even Fibonacci number, which forces $|F| = 2$ because even numbers repeat every third term. So $n$ has exactly $5$, impossible because $n$ has $2^k$ divisors. So this case yields no solutions. 2) $1 \in F$ In this case, for some positive integer $u \geqslant 2$ we know that all fibonacci numbers $1, 2, 3, \dots, F_u$ divide $n$. Observe that $8 \not \mid 8$ as $n$ is squarefree. Thus $u = 2, 3, 4$. If $u = 2$, then only $1, 2$ are in $F$. Then $2^k = |F|+|T|+1 \geqslant 5$ which implies that $n$ is of the form $2pq$. But then one of $p, q$ exceeds $5$ which cannot be triangular. Thus $u = 3$ or $4$. If $u = 3$ we have $F = \{1, 2, 3\}$. Because no prime greater than $3$ divides $n$ (it can't be in $F$ or $T$) we have $n = 2\cdot 3 = 6$ which does not work because it has only $3$ proper divisors. Thus $u = 4$ and we have $F = \{1, 2, 3, 5\}$. This implies that all of $2, 3, 5 \mid n$ and so $30 \mid n$. Thus $6, 10, 15$ are all proper divisors of $n$ and these must be in $T$. Observe that no prime $p > 5$ can divide $n$ and $2^k$ must be at least $4+3+1 = 8$ so $n = 2 \cdot 3 \cdot 5$ which indeed works with $F = \{1, 2, 3, 5\}$ and $T = \{6, 10, 15\}$. To conclude the only such positive integer $n$ is $n = 30$.