Let
$$
G_n:=\max\{length(S), \ \text{such that $S$ is a solid sequence and } S\subset\{1,\dots,n\}\}.
$$
Then one has $G_1=1$, $G_2=3$, $G_3=7$.
Your claim is that $G_{2^{m-1}-1}<2^m$. This is false. The example below shows that
$$
G_{F_n}\ge 2^{n-1}-1,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad (Ex)
$$
where $F_n$ is the $n$th term in the Fibonacci sequence. In particular,
for $m=5$, we have $15=2^{m-1}-1>F_7=13$, and so
$$
G_{15}=G_{2^{m-1}-1}\ge G_{F_7}\ge 2^{6}-1>2^m=32.
$$
In particular, the solid sequence of length $63$ with no term higher than $15$ is
$$
(1,13,8,13,5,13,8,13,3,13,8,13,5,13,8,13,2,13,8,13,5,13,8,13,3,13,8,13,5,13,8,13,
$$
$$
1,13,8,13,5,13,8,13,3,13,8,13,5,13,8,13,2,13,8,13,5,13,8,13,3,13,8,13,5,13,8),
$$
Note that $(Ex)$ provides a lower bound (as will do any family of examples), instead of the desired upper bound.
I suspect that this lower bound is also an asymptotic upper bound, but I have no proof.
$\bf{The\ Example}:$
For each $n\ge 1$ we will provide inductively an example of a sequence $S^{(n)}$ with $length(S^{(n)})=2^{n}$, such that
$(S^{(n)}_1,\dots,S^{(n)}_{2^n-1})$ is a solid sequence, and such that $S^{(n)}\subset\{1,\dots,F_{n+1}\}$.
First terms:
For $n=1$ clearly $S^{(1)}=(1,1)$ satisfies the hypothesis.
For $n=2$ the sequence is $S^{(2)}=(1,2,1,2)$.
For $n=3$ we have $S^{(3)}=(1,3,2,3,1,3,2,3)$ and
for $n=4$ we have $S^{(4)}=(1,5,3,5,2,5,3,5,1,5,3,5,2,5,3,5)$.
The inductive construction is the following:
We set
$$
S^{(n)}_{2k}:=F_{n+1}\quad\text{and}\quad S^{(n)}_{2k-1}:=S^{(n-1)}_{k}\quad\text{for $k=1,\dots, 2^{n-1}$}.
$$
Now we prove inductively that $(S^{(n)}_1,\dots,S^{(n)}_{2^n-1})$ is a solid sequence.
For this we prove a little more:
(1.) For each $S=S^{(n)}$ and any two consecutive sets $A,B\subset S^{(n)}$, if $\#A>\#B$, then $\sum_{i\in A}S_i> \sum_{i\in B}S_i$.
(2.) For each $S=S^{(n)}$ and any two consecutive sets $A,B\subset S^{(n)}$, if $\#A=\#B+1$, then $\sum_{i\in A}S_i- \sum_{i\in B}S_i<F_{n+2}$.
(3.) For each $S=S^{(n)}$ and any two consecutive sets $A,B\subset S^{(n)}$, if $\#A=\#B$, then $|\sum_{i\in A}S_i- \sum_{i\in B}S_i|<F_{n+1}$.
(4.) For each $S=S^{(n)}$ and any two consecutive sets $A,B\subset S^{(n)}$, if $\#A=\#B$, then $\sum_{i\in A}S_i\ne \sum_{i\in B}S_i$, except when
$\#A=\#B=2^{n-1}$ (and so $A\cup B=S^{(n)}$).
Here we consider at the same time the two cases in which $A$ lies above $B$ and viceversa.
For $n=1,2,3$ one can verify (1.)--(4.) by hand.
Assume (1.), (2.), (3.), (4.) are true for $n-1$.
We first prove (3.) and (4.).
If $\#A=\#B =2k$, then (3.) and (4.) are clear form the inductive hypothesis, since both sets have the same number of $F_{n+1}$'s. Once you remove them,
by inductive hypothesis the sums satisfy the required conditions (one has to be careful with the case $A\cup B= S^{(n)}$).
If $\#A=\#B =2k-1$, then one of them has one term more with value $F_{n+1}$, and the other has one term more of $S^{(n-1)}$.
Assume $A$ has one term more with value $F_{n+1}$ and write $A_{n-1}$ and $B_{n-1}$ for the sets $A$ and $B$ with all the terms $F_{n+1}$ removed.
Then
$$
0<\sum_{i\in A}S^{(n)}_i-\sum_{i\in B}S^{(n)}_i=F_{n+1}-(\sum_{i\in B_{n-1}}S^{(n-1)}_i-\sum_{i\in A_{n-1}}S^{(n-1)}_i)<F_{n+1},
$$
which proves (3.) and (4.) in this case.
In fact, the equality is clear, and the last inequality follows from the inductive hypothesis item (1.), since $\#A_{n-1}<\#B_{n-1}$. The first inequality follows from the inductive hypothesis item (2.), since $\#B_{n-1}=\#A_{n-1}+1$, and so
$$
\sum_{i\in B_{n-1}}S^{(n-1)}_i-\sum_{i\in A_{n-1}}S^{(n-1)}_i<F_{(n-1)+2}=F_{n+1}.
$$
Now we prove (1.).
Assume $\#A>\#B$ and define $A_{n-1}$ and $B_{n-1}$ as before. Then
$$
\sum_{i\in A}S_i-\sum_{i\in B}S_i=F_{n+1}(\#\{F_{n+1}\ \text{ in }A\}-\#\{F_{n+1}\ \text{ in }B\})+ \sum_{i\in A_{n-1}}S^{(n-1)}_i-\sum_{i\in B_{n-1}}S^{(n-1)}_i
$$
and there are two cases:
a) $\#A_{n-1}>\# B_{n-1}$ and $\#\{F_{n+1}\ \text{ in }A\}$ is greater than or equal to $\#\{F_{n+1}\ \text{ in }B\}$. In this case clearly
$\sum_{i\in A_{n-1}}S^{(n-1)}_i-\sum_{i\in B_{n-1}}S^{(n-1)}_i>0$ by inductive hypothesis item (1.) and so item (1.) for $n$ follows directly.
b) $\#A_{n-1}=\# B_{n-1}$ and $\#\{F_{n+1}\ \text{ in }A\}$ is greater than $\#\{F_{n+1}\ \text{ in }B\}$. In this case, if
$\sum_{i\in A_{n-1}}S^{(n-1)}_i-\sum_{i\in B_{n-1}}S^{(n-1)}_i>0$, then item (1.) for $n$ follows directly.
Else $0<\sum_{i\in B_{n-1}}S^{(n-1)}_i-\sum_{i\in A_{n-1}}S^{(n-1)}_i<F_n$ by inductive hypothesis item (3.), and so
$$
\sum_{i\in A}S_i-\sum_{i\in B}S_i\ge F_{n+1}-\left( \sum_{i\in B_{n-1}}S^{(n-1)}_i-\sum_{i\in A_{n-1}}S^{(n-1)}_i\right)>F_{n+1}-F_n>0.
$$
Finally we prove (2.).
Assume $\#A=\#B+1$ and define $A_{n-1}$ and $B_{n-1}$ as before. Then
$$
\sum_{i\in A}S_i-\sum_{i\in B}S_i=F_{n+1}(\#\{F_{n+1}\ \text{ in }A\}-\#\{F_{n+1}\ \text{ in }B\})+ \sum_{i\in A_{n-1}}S^{(n-1)}_i-\sum_{i\in B_{n-1}}S^{(n-1)}_i
$$
and there are two cases:
a) $\#A_{n-1}>\# B_{n-1}$ and $\#\{F_{n+1}\ \text{ in }A\}$ is equal to $\#\{F_{n+1}\ \text{ in }B\}$. In this case clearly
$\sum_{i\in A_{n-1}}S^{(n-1)}_i-\sum_{i\in B_{n-1}}S^{(n-1)}_i<F_{n+1}<F_{n+2}$ by inductive hypothesis item (2.) and so item (2.) for $n$ follows directly.
b) $\#A_{n-1}=\# B_{n-1}$ and $\#\{F_{n+1}\ \text{ in }A\}$ is $\#\{F_{n+1}\ \text{ in }B\}$ plus one. In this case,
by inductive hypothesis item (3.) we have
$$
\sum_{i\in A_{n-1}}S^{(n-1)}_i-\sum_{i\in B_{n-1}}S^{(n-1)}_i< F_n
$$
and so
$$
\sum_{i\in A}S_i-\sum_{i\in B}S_i = F_{n+1}+\left( \sum_{i\in A_{n-1}}S^{(n-1)}_i-\sum_{i\in B_{n-1}}S^{(n-1)}_i\right)<F_{n+1}+F_n=F_{n+2},
$$
as desired.
Clearly $(1.)$ and $(4.)$ together imply that $(S^{(n)}_1,\dots,S^{(n)}_{2^n-1})$ is a solid sequence.
