Your RHS can be factorized as $2k(4k-1)(8k^2-4k+1)$.
please take note of my previous post as follows :
$ 8k^2-4k+2=8k^2-2k -------------------(*) $Albert said:
But how did you get $x=8k^2−4k+1$ then?
What this shows is that if the integer $N(k) \overset{\text{def}}{=} 64k^4 - 48k^3 + 16k^2 -2k$ is factorised as the product of the two factors $8k^2-2k$ and $8k^2-4k+1$, then the only way for those factors to differ by $1$ is if $k=1$. That still leaves open the possibility that $N(k)$ could be factorised in some different way as the product of two integers differing by $1$. So I agree with ILikeSerena that the above argument does not rule out the possibility of other solutions (though admittedly it seems unlikely that there are any).
I checked up to k=100000 with N(100000) = 6399952000159999800000.
That would only hold if one of those factors is a prime number.
I think that you are confusing algebraic factorisation with arithmetic factorisation. To take a simple example, if $f(x) = x^2+6x+5$ then $f(x)$ has only one algebraic factorisation, namely $f(x) = (x+1)(x+5)$. If you put $x=1$, then $f(1) = 12$ and the two factors $x+1$ and $x+5$ give you the factorisation $f(1) = 2\times 6$. But if you consider it arithmetically, as a number, then $12$ has other factorisations, such as $12 = 3\times 4$, which are not apparent from the algebraic factorisation of $f(x)$.