Below are $3$ proofs of the gcd distributive law $\rm\:(ax,bx) = (a,b)x\:$ using Bezout's identity, universal gcd laws, and unique factorization.
First we show that the gcd distributive law follows immediately from the fact that, by Bezout, the gcd may be specified by linear equations. Distributivity follows because such linear equations are preserved by scalings. Namely, for naturals $\rm\:a,b,c,x \ne 0$
$\rm\qquad\qquad \phantom{ \iff }\ \ \ \:\! c = (a,b) $
$\rm\qquad\qquad \iff\ \: c\:\ |\ \:a,\:b\ \ \ \ \ \ \&\ \ \ \ c\ =\ na\: +\: kb,\ \ \ $ some $\rm\:n,k\in \mathbb Z$
$\rm\qquad\qquad \iff\ cx\ |\ ax,bx\ \ \ \&\ \ \ cx = nax + kbx,\,\ \ $ some $\rm\:n,k\in \mathbb Z$
$\rm\qquad\qquad { \iff }\ \ cx = (ax,bx) $
The reader familiar with ideals will note that these equivalences are captured more concisely in the distributive law for ideal multiplication $\rm\:(a,b)(x) = (ax,bx),\:$ when interpreted in a PID or Bezout domain, where the ideal $\rm\:(a,b) = (c)\iff c = gcd(a,b)$
Alternatively, more generally, in any integral domain $\rm\:D\:$ we have
Theorem $\rm\ \ (a,b)\ =\ (ax,bx)/x\ \ $ if $\rm\ (ax,bx)\ $ exists in $\rm\:D.$
Proof $\rm\quad\: c\ |\ a,b \iff cx\ |\ ax,bx \iff cx\ |\ (ax,bx) \iff c\ |\ (ax,bx)/x\ \ \ $ QED
The above proof uses the universal definitions of GCD, LCM, which often served to simplify proofs, e.g. see this proof of the GCD * LCM law.
Alternatively, comparing powers of primes in unique factorizations, it reduces to the following
$$\begin{eqnarray} \min(a+x,\,b+x) &\,=\,& \min(a,b) + x\\
\rm expt\ analog\ of\ \ \ \gcd(a \,* x,\,b \,* x)&=&\rm \gcd(a,b)\,*x\end{eqnarray}\qquad\qquad\ \ $$
The proof is precisely the same as the prior proof, replacing gcd by min, and divides by $\,\le,\,$ and
$$\begin{eqnarray} {\rm employing}\quad\ c\le a,b&\iff& c\le \min(a,b)\\
\rm the\ analog\ of\quad\ c\ \, |\, \ a,b&\iff&\rm c\ \,|\,\ \gcd(a,b) \end{eqnarray}$$
$\ c \le a,b \!\iff\! c\!+\!x \le a\!+\!x,b\!+\!x\!\iff\! c\!+\!x \le \lfloor a\!+\!x,b\!+\!x\rfloor\!\iff\! c \le \lfloor a\!+\!x,b\!+\!x\rfloor \!-\!x$
where $\,\lfloor y,z\rfloor := \min(y,z).$
If an odd prime $p$ divides $2^n+1$, then the order of $2$ modulo $p$ is even (it is a divisor of $2n$, but not of $n$). If an odd prime $q$ divides $2^m-1$ with $m$ odd, then the order of $2$ modulo $q$ is odd (it is a divisor of $m$). Hence $p \neq q$. Since $2^m - 1$ is odd for $m > 0$, in particular all odd $m$, the greatest common divisor cannot be even. So no prime divides both, $2^n+1$ and $2^m-1$.
Alternatively, we can use
$$\gcd (2^t-1, 2^u-1) = 2^{\gcd (t,u)}-1\tag{1}$$
to conclude
$$\gcd (2^m-1, 2^{2n}-1) = 2^{\gcd(m,2n)}-1.$$
But since $m$ is odd, we have $\gcd (m,2n) = \gcd(m,n)$, and hence
$$2^{\gcd(m,2n)}-1 \mid 2^n-1,$$
which, since
$$\gcd(2^n-1,2^n+1) = \gcd(2^n-1,2) \mid 2$$
and $2^{\gcd(m,2n)}-1$ is odd, implies $\gcd (2^{\gcd(m,2n)}-1,2^n+1) = 1$ and hence $\gcd(2^m-1,2^n+1) = 1$.
To see $(1)$, write $u = q\cdot t + r$ with $0 \leqslant r < t$, and
$$2^u-1 = 2^r\left(2^{q\cdot t}-1\right) + \left(2^r-1\right),$$
which, since $2^t-1 \mid (2^t)^q-1$, yields
$$\gcd(2^t-1,2^u-1) = \gcd(2^t-1,2^r-1),$$
and continuing the Euclidean algorithm for the exponents finally yields $(1)$.
Best Answer
Well $k= 2m+1$ is odd so $k^2 - 1= 4m^2 + 4m$ so $\gcd(8,4m^2+4m) = 4\gcd(2,m^2+m)= 4\gcd(2, m(m+1))$.
And $\gcd(2,m(m+1))$ is $2$ if $m(m+1)$ is even and $1$ if $m(m+1)$ is odd.
And either $m$ is even and $m(m+1)$ is even; or $m$ is odd and $m+1$ is even and $m(m+1)$ is even. SO $m(m+1)$ is even.
So $\gcd(8,k^2 -1)=\gcd(8,4m^2 +4m) = 4\gcd(2,m(m+1))=4*2 =8$.
The only real trouble is there's nothing there that couldn't have been explained (and probably easierly) without gcd.