Why does fusion stop at iron when nickel is most tightly bound?

The iron-peak elements are mostly the product of alpha capture reactions onto nuclei that begin with a similar number of neutrons and protons ($Z = N$).

The nuclear burning associated with carbon and oxygen (in type Ia supernovae) or silicon (in the cores of massive stars at the ends of their lives) is very fast or even explosive. The important reactions in determining the immediate final products are those that proceed on fast timescales.

In these cases, there is a competition between alpha capture and photodisintegration with the constraint that $Z \simeq N$, since weak, flavour-changing reactions are generally too slow to move the neutron/proton ratio far from unity before an equilibrium has been reached between alpha capture and photodisintegration.

Subject to these constraints, then 56Ni turns out to be the most stable nucleus. Further alpha captures to 60Zn (or beyond) are not favored$^*$, because the higher temperatures that would be required to overcome the greater Coulomb barrier results in photodisintegration to smaller nuclei.

Hence there is no easy route to 62Ni.

The fact that 56Fe dominates the iron-peak element abundances that are seen in the atmospheres of stars and the interstellar medium is because the 56Ni in supernovae ejecta decays to 56Co and then 56Fe on half-lives of 6 and 77 days respectively. Inside the dense core of a massive star then electron capture by 56Ni can be more rapid, but still results in 56Fe.

Edit:

$*$ Note that alpha capture onto 56Ni is still energetically favourable in isolation. However, in a core made of 56Ni, then the alpha particle has to be stripped off a 56Ni nucleus first, and the net rearrangement of nucleons would be endothermic.

$^{56}Ni$ is produced in silicon-fusion stars. The fusion process doesn't "stop" at $Fe$. Several A=56 nuclides show up. See the Wiki-pedia article on :Silicon burning.

Also, Introductory Nuclear Physics by Krane, Chapter 19, Section 4.