Here is an alternate (and complete) proof of 2. Following Mark Sapir, we show that the only solution of $$ \frac{2^{tm}-1}{2^m-1} = 3^r $$ in positive integer triplets $(t,m,r)$ is $(2,1,1)$ and $(2,3,2)$. I am sure this was written down before, by the way.
Our main tool is the observation that $\mathrm{ord}_{3^n}(4)=3^{n-1}$ when $n\geq 1$ (in other words $2$ is a primitive root modulo $3^n$), which implies that $v_3(4^k-1)=1+v_3(k)$. The observation follows from $4^{3^{n-2}}\equiv 1+3^{n-1}\pmod{3^n}$ when $n\geq 2$, which in turn can be proved by induction or by the binomial theorem.
In the diophantine equation $r\geq 1$, hence $tm$ is even. If $m$ is even, then $$ r=v_3\left (\frac{2^{tm}-1}{2^m-1}\right)=(1+v_3(tm))-(1+v_3(m))=v_3(t), $$ so that $t$ is divisible by $3^r$. This is a contradiction: $$ 3^r=\frac{2^{tm}-1}{2^m-1} > t \geq 3^r, $$ which proves that $m$ is odd and $t$ is even. In that case $4^m-1\mid 2^{tm}-1$, and $$ \frac{2^{tm}-1}{4^m-1} = 3^s $$ for some $0\leq s\leq r$. As before $$ s=v_3\left (\frac{2^{tm}-1}{4^m-1}\right)=(1+v_3(tm))-(1+v_3(m))=v_3(t), $$ so that $t$ is divisible by $3^s$. This is a contradiction when $t\geq 4$: $$ 3^s=\frac{2^{tm}-1}{4^m-1} > 4(t/2-1)\geq t \geq 3^s, $$ which proves that $t=2$. Now the original diophantine equation becomes $$ 2^m+1=3^r,$$ where $m$ is odd. Then $1+v_3(m)=v_3(4^m-1)=r$, so that $m$ is divisible by $3^{r-1}$. It follows that $1+2^{3^{r-1}}\leq 3^r$. It is easy to see that this holds only for $r=1$ and $r=2$, which yields $m=1$ and $m=3$, respectively.