Let us begin. The first thing I notice is that \(\sigma(n)\) is bad, but also that \(\tau(n)\) must be odd - so \(n\) is a perfect square. Let \[n = p_1^{2a_1} p_2^{2a_2} \dots p_k^{2a_k}\] Now, \(\tau(n) = (2a_1 + 1)(2a_2 + 1)\dots (2a_k + 1)\). Choose a prime \(q \mid \tau(n)\), so that we have \[2^{\sigma(n) - 1} \equiv -1 \pmod{q} \implies 2^{2(\sigma(n) - 1)} \equiv 1 \pmod{q}\] \[\sigma(n) = \frac{p_1^{2a_1 + 1} - 1}{p_1 - 1} \frac{p_2^{2a_2 + 1} - 1}{p_2 - 1} \dots \frac{p_k^{2a_k + 1} - 1}{p_k - 1}\] Note that \(\sigma(n)\) is odd. Let \(r\) be the order of \(2\) modulo \(q\), which then means that \(r \mid \gcd(q - 1, 2(\sigma(n) - 1))\). But anyways, what is this weird \(\sigma(n) - 1\)? How are we supposed to deal with it?

The best idea when lost is to use small cases. Number theory is combinatorics on primes, so suppose \(n = p^{2a}\). Then \(\tau(n) = 2a + 1\), and \(\sigma(n) = \frac{p^{2a + 1} - 1}{p - 1}\). Now we again choose \(q \mid 2a + 1\), and note that \[q \mid 2^{\frac{p^{2a + 1} - 1}{p - 1} - 1} + 1 \implies 2^{2\left(\frac{p^{2a + 1} - 1}{p - 1} - 1\right)} \equiv 1 \pmod{q}\] and we again introduce \(r\), to get \(r \mid \gcd(p - 1, 2(\sigma(n) - 1))\). The prime \(p\) is basically any random odd prime, and we want to show that it's just not possible for the above congruence to hold. Of course, we should note that \(2a + 1\) is really just \(qk\) for some odd \(k\), and we want the impossibility of \[2^{2\left(\frac{p^{qk} - 1}{p - 1} - 1\right)} \equiv 1 \pmod{q}\] for every odd prime \(p\) and odd \(k\). The only way for this to be impossible is if the exponent can never be a multiple of \(r\), which is fixed for a fixed \(q\). I mean, the problem is true, which is why we're concluding these things. The fact that this is true is quite frankly insane - the only thing we know about \(r\) for certain is that \(r \mid q - 1\), in fact, \(r = q - 1\) is also possible when \(2\) is a primitive root modulo \(q\). Okay, I've figured it out! Consider any odd prime exponent \(s^b\) that divides \(r\). Then, \[s^b \mid \frac{p^{qk} - 1}{p - 1} - 1\] is easy to rig by choosing some particular \(p\) such that \(s \nmid p - 1\), and \(qk \equiv 1 \pmod{s^b - s^{b-1}}\). For multiple \(s\), we don't need to CRT, we can just take the product \(m \prod (s^b - s^{b-1}) + 1\), and \(q\) does not interfere because \(s \le r \le q - 1 < q\). So the only boundary between our brains and the problem being false is the prime exponent \(2^d\), and apparently \[2^d \mid 2 \left(\frac{p^{qk} - 1}{p - 1} - 1 \right) \iff 2^{d-1} \mid \frac{p^{qk} - 1}{p - 1} - 1\] is impossible to achieve. Remember that \(2^d \mid r \mid q - 1\). But how do you calculate the \(\nu_2\) of the above expression, and that too in terms of \(d\)? Of course, the \(-1\) is super annoying, so upon removing it, we are met with a better expression: \[\frac{p^{qk} - 1}{p - 1} - 1 = \frac{p(p^{qk-1} - 1)}{p-1}\] Calculating the \(\nu_2\) of this is thankfully feasible, somewhat, since \[\nu_2\left(\frac{p(p^{qk-1} - 1)}{p-1} \right) = \nu_2(p^{qk-1} - 1) - \nu_2(p - 1) = \nu_2(p + 1) + \nu_2(qk - 1) - 1\] So it's definitely possible to get this expression to be \(\ge d - 1\) simply by varying \(p\). What is the problem then? We must have missed something. Indeed, note that \(r\) does not even divide the LHS, \(r\) only divides \(2\) times (!) the LHS because the problem explicitly states \(\tau(n) \mid 2^{\sigma(n) - 1} + 1\), rather than the \(2 ( \sigma(n) - 1)\) version. Now we are using (almost) all the information we have, and we want (the impossibility of) \[d = \nu_2(p+ 1) + \nu_2(qk - 1)\] This is just not dependent on \(p\), which can vary as it wants to, which means \(qk - 1\) must be the real problem. In order to actually use all the information we have, we have to remember that \(2^d \mid q - 1\). Since we know what the contradiction we want is, it is easy to see (contrary to when we have no motivation at all) that \(q' \equiv 1 \pmod{2^{\nu_2(\sigma(n) - 1) + 1}}\) is a universal fact, true for all factors of \(qk = 2a + 1\). Thus, \(qk \equiv 1 \pmod{2^d}\), \(qk - 1 \equiv 0 \pmod{2^d}\), and \[d = \nu_2(p+ 1) + \nu_2(qk - 1) \ge 1 + d\] a contradiction!

Finally, we have solved the small case. As is tradition, the general case should be trivial now. Let \(q \mid \tau(n)\) be some prime, and note that if \(r\) is the order of \(2\) modulo \(q\), then \(r \mid \gcd(q - 1, 2 (\sigma(n) - 1))\). Since \(r \nmid \sigma(n) - 1\), we must have \(\nu_2(\sigma(n) - 1) = d -1\), \(\nu_2(r) = d\), and \(\nu_2(q - 1) \ge d\). Because \(\nu_2(\sigma(n) - 1)\) is a constant regardless of what divisor of \(\tau(n)\) we choose, we must have \(\nu_2(q - 1) \ge d\) for all \(q \mid \tau(n)\). Thus, \[q \equiv 1 \pmod{2^d} \implies 2a_i + 1 \equiv 1 \pmod{2^d} \implies 2a_1 \equiv 0 \pmod{2^d} \] Now using LTE on individual terms, we obtain similar to above that \[\nu_2\left(\frac{p^{qk} - 1}{p - 1} - 1 \right) \ge d \implies \frac{p^{qk} - 1}{p - 1} \equiv 1 \pmod{2^d}\] Hence, taking the product, \[\sigma(n) \equiv 1 \pmod{2^d} \implies \nu_2(\sigma(n) - 1) = d \neq d - 1\] and we are done.