This is a partial answer.
This answer proves the following three claims :
Claim 1 : $\binom n5=2^r-1$ has no non-trivial solutions.
Claim 2 : $\binom n7=2^r-1$ has no non-trivial solutions.
Claim 3 : $\binom n9=2^r-1$ has no non-trivial solutions.
Let us consider $$\binom nm=2^r-1\tag1$$ where $m$ is odd.
The equation $(1)$ can be written as
$$2^rm!=n(n-1)\cdots (n-m+1)+m!\tag2$$
We can see that if $m$ is odd, then the RHS is divisible by $n+1$.
Letting $N:=n+1$, $(2)$ can be written as
$$2^rm!=(N-1)(N-2)\cdots (N-m)+m!\tag3$$
whose RHS is divisible by $N$ when $m$ is odd.
For $m=3$, Benjamin Wright has shown that there are no non-trivial solutions.
Claim 1 : $\binom n5=2^r-1$ has no non-trivial solutions.
Proof :
For $m=5$, $(3)$ can be written as
$$2^{r+3}\cdot 3\cdot 5=N\times f_5(N)$$
where $f_5(N)=N^4 - 15N^3+ 85 N^2 - 225 N + 274$.
Here, we have the followings :
$N\ge 8$
$N$ is not divisible by $4$
(Proof : We have $f_5'(N)=(N - 5) (4(N-8)^2+39(N-8)+101)\gt 0$, so $f_5(N)\ge f_5(8)=330$. Suppose that $N$ is divisible by $4$. Then, $f_5(N)$ is of the form $2\times (\text{odd})$, so $f_5(N)\le 2\cdot 3\cdot 5=30$ which contradicts $f_5(N)\ge 330$)
$N\not\equiv 3\pmod 4$
(Proof : Suppose that $N\equiv 3\pmod 4$. Then, $n(n-1)\cdots (n-4)$ is divisible by $2^4$, so $\binom n5$ is even where $5!$ is not divisible by $2^4$)
$N\not\equiv 1,2,3,4,5\pmod 8$
(Proof : Suppose that $N\equiv 1,2,3,4,5\pmod 8$. Then, $n(n-1)\cdots (n-4)$ is divisible by $2^4$, so $\binom n5$ is even)
So, it is necessary that $N=30$, i.e. $n=29$ which is not sufficient.
Therefore, there are no non-trivial solutions.$\ \blacksquare$
Claim 2 : $\binom n7=2^r-1$ has no non-trivial solutions.
Proof :
For $m=7$, $(3)$ can be written as
$$2^{r+4}\cdot 3^2\cdot 5\cdot 7=N\times f_7(N)$$
where $f_7(N)=N^6 - 28 N^5 + 322 N^4 - 1960 N^3 + 6769 N^2 - 13132 N + 13068$.
Here, we have the followings :
$N\ge 10$
$N\not\equiv 0\pmod 9$
(Proof : $f_7(N)$ is always divisible by $3$ since $f_7(N)\equiv N (N+1)^2 (N-1)^3\pmod 3$. Suppose that $N\equiv 0\pmod 9$. Then, $Nf_7(N)$ is divisible by $3^3$, a contradiction)
$N$ is not divisible by $8$
(Proof : We have $f_7'(N)=2 (N - 7) (3(N-10)^4+71(N-10)^3+631(N-10)^2+2487(N-10)+3708)\gt 0$, $f_7(N)\ge f_7(10)=18648$. Suppose that $N$ is divisible by $8$. Then, $f_7(N)$ is of the form $4\times (\text{odd})$, so $f_7(N)\le 2^2\cdot 3^2\cdot 5\cdot 7=1260$ which contradicts $f_7(N)\ge 18648$)
$N\not\equiv 1,2,3\pmod 4$
(Proof : Suppose that $N\equiv 1,2,3\pmod 4$. Then, $n(n-1)\cdots (n-6)$ is divisible by $2^5$, so $\binom n7$ is even where $7!$ is not divisible by $2^5$)
$N\not\equiv 4\pmod 8$
(Proof : Suppose that $N\equiv 4\pmod 8$. Then, $n(n-1)\cdots (n-6)$ is divisible by $2^5$, so $\binom n7$ is even)
Therefore, there are no non-trivial solutions.$\ \blacksquare$
Claim 3 : $\binom n9=2^r-1$ has no non-trivial solutions.
Proof :
For $m=9$, $(3)$ can be written as
$$2^{r+7}\cdot 3^4\cdot 5\cdot 7=N\times f_9(N)$$
where $f_9(N)=N^8 - 45 N^7 + 870 N^6 - 9450 N^5 + 63273 N^4 - 269325 N^3 + 723680 N^2 - 1172700 N + 1026576$.
Here, we have the followings :
$N\ge 12$
$N\not\equiv 0\pmod{27}$
(Proof : $f_9(N)$ is always divisible by $9$ since $f_9(N)\equiv N^2 (N-1)^3(N+1)^3\pmod 9$. Suppose that $N\equiv 0\pmod{27}$. Then, $Nf_9(N)$ is divisible by $3^5$, a contradiction)
$N$ is not divisible by $2^5$
(Proof : We have $f_9'(N)=(N-9)(8(N-12)^6+333(N-12)^5+5733(N-12)^4+52191(N-12)^3+264999(N-12)^2+712116(N-12)+795580)\gt 0$, so $f_9(N)\ge f_9(12)=1693440$. Suppose that $N$ is divisible by $2^5$. Then, $f_9(N)$ is of the form $2^4\times (\text{odd})$, so $f_9(N)\le 2^4\cdot 3^4\cdot 5\cdot 7=45360$ which contradicts $f_9(N)\ge 1693440$)
$N$ is even
(Proof : Suppose that $N$ is odd. Then, since $n$ is even, $n(n-1)\cdots (n-8)$ is divisible by $2^8$, so $\binom n9$ is even where $9!$ is not divisible by $2^8$.)
$N\not\equiv 2,4,6,8\pmod{16}$
(Proof : Suppose that $N\equiv 2,4,6,8\pmod{16}$. Then, $n(n-1)\cdots (n-8)$ is divisible by $2^8$, so $\binom n9$ is even)
So, it is necessary that $N=12,14,16,28,30,42,48,60,80,90,112,126,140,144,240,252,336,560,720,1008,1260,1680,5040$.
So, it is necessary that $n=11,13,15,27,29,41,47,59,79,89,111,125,139,143,239,251,335,559,719,1007,1259,1679,5039$ none of which is sufficient.
Therefore, there are no non-trivial solutions.$\ \blacksquare$
