(This proof was completed with an insightful comment from Gottfried. I'm placing it as an answer so that it is readily seen that an answer exists, as opposed to just leaving it in the question.)
Suppose we have $ 2^n \pm 1 = x^k$ for some positive integers $x, k$. Cases $n=1, 2, 3$ are easily dealt with. $n=3$ yields the solution $2^3 + 1 = 3^2$. Henceforth, let $n\geq4$. Furthermore, a simple check shows that $x\neq 1, 2$.
First, we will deal with the power $k=2l$ being even. Rewriting $x^{2l}$ as $(x^l)^2$, we may assume that $k=2$.
Proof: $2^n -1 \equiv 3 \pmod{4}$ hence is never a square.
If $2^n +1 =x^2$, then $2^n = (x-1)(x+1)$ and both of these are powers of 2 that differ by 2. Thus we must have $(x-1) = 2, (x+1) = 4$, which gives $2^n = 8$ so $n=3$ (reject). $_\square$
Now, we deal with $k$ odd.
Proof: Say $2^n +1 = x^k$. Then $2^n = x^k - 1 = (x-1)(x^{k-1}+x^{k-2} + \ldots +1)$, so both terms are powers of 2. We have $ x -1$ is a power of 2 greater than 1, hence is even. Thus $x$ is odd. But the other term is the sum of $k$ odd numbers, hence is odd. Since this is clearly greater than 1, we get a contradiction.
Say $2^n -1 = x^k$. Then $2^n = x^k + 1 = (x+1)(x^{k-1} - x^{k-2} + \ldots -x +1 )$, so both terms are powers of 2. We have $x+1$ is a power of 2 that is greater than 1, hence is even. Thus $x$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since $x^p + 1 \neq x+1$ except for $p=1$, this means that the term is greater than 1. Hence we get a contradiction. $ _\square$
In general, the Diophantine equation
$$
n!+k=m^2
$$
is "easy" for $k$ being a non-square integer and hard for $k$ being a perfect square. In fact, if $k$ is not a square it follows that $n\le k$. In this case, for small $k$, we only have to check a few possible values for $n$. The most famous case, where $k$ is a perfect square is the Brocard equation $n!+1=m^2$, which is unsolved until today.
For a reference see the following post with very informative answers:
For any $k \gt 1$, if $n!+k$ is a square then will $n \le k$ always be true?
Best Answer
Suppose that $63 \cdot 2^n + 1=a^b$, with $b \geq 2$, $a$ not being a power and $n \geq 8$. Then $a$ is odd.
Assume that $b$ is odd. Then $n=v_2(a^b-1)=v_2(a-1)$, so $a-1 \geq 2^n$ and $63\cdot 2^n=a^b-1 \geq (a-1)^b \geq 2^{nb}$, so that $256^{b-1} \leq 2^{n(b-1)} \leq 63$, hence $b=1$. We get a contradiction.
Assume that $63 \cdot 2^n-1=a^b$ with $b$ odd and $n \geq 6$. Then $n=v_2(a^b+1)=v_2(a+1)$, so $a \geq 2^{n}-1$. Then $2^{n+6} \geq 63 \cdot 2^n = a^b+1 \geq (2^n-1)^b + 1 \geq 2^{b(n-1)}$, so $b(n-1) < n+6 \leq 2(n-1)$, and $b=1$. We get a contradiction.
So by your remark, we must ask when $63 \cdot 2^n=a^2-1$ for some odd $a$ with $n \geq 8$.
One of $a-1$ and $a+1$ is not divisible by $4$ but divides $63 \cdot 2^n$, so $a-1 \leq 63 \cdot 2=126$. Thus $63 \cdot 2^n=a^2-1 \leq 126 \cdot 128$, hence $n \leq 8$.