Is there some sophisticated method (or maybe some easy one, though I doubt it) to show that the only solution to $m^2 = 14k^4 - 6k^2 + 1$ in positive integers is $k=1$, $m=3$? Perhaps something around elliptic curves and SAGE computation or whatever seems fine, though the simplest attack will be most appreciated!
When trying this out by hand, the best idea I had what to write it like $m^2 - 14(k^2 - \mbox{something})^2 = \mbox{something}$ and then consider this Pell-type equation, but the presence of $k^2$ in one of the Pell variables makes things much harder.
$$m^2=14k^4-6k^2+1\tag{1}$$
An equation $(1)$ is birationally equivalent to the elliptic curve $$Y^2 = X^3 - 68X + 208\tag{2}$$
with $$k = \frac{2X-8}{Y}, m = \frac{X^3-12X^2-144+68X}{Y^2}$$
According to LMFDB , elliptic curve has seven integer points as follows.
$(-6,\pm 20)$, $ \left(4, 0\right) $, $(6,\pm 4)$, $(13,\pm 39).$
Hence we get the positive integer solution $(k,m)=(1,3)$ from $(X,Y)=(6,\pm 4).$