The Diophantine equation $p^x+6^y=z^{2n}$: a complete solution for half of the primes

Let $p$ be a prime number and let $n\ge 1$ be an integer. We completely solve the exponential Diophantine equation $p^x+6^y=z^{2n}$ in positive integers $x,y,z$ for every prime $p$ such that the Legendre symbol $\text{leg}6p$ equals $-1$, that is, for $p\equiv 7,11,13,17(\mathrm{m}\mathrm{o}\mathrm{d}24)$: the equation has no solution, with the single exception $(p,n,x,y,z)=(13,1,1,2,7)$. Since these residue classes contain half of all primes in the sense of Dirichlet density, this settles the equation for one prime out of two, uniformly in the exponent $n$. Two ingredients of the proof have independent interest. First, we show that the Ramanujan–Nagell type equation $p^x=2\cdot 6^m+1$ has no solution with $x\ge 2$, using Zsigmondy’s theorem on primitive prime divisors; this removes the extra congruence hypotheses that appeared in earlier work of the author on the case $n=1$ and solves the open problems formulated there. Secondly, we prove that ${13}^x+6^y=z^2$ has $(x,y,z)=(1,2,7)$ as its unique solution, thereby completely resolving the exceptional prime. The Legendre condition is sharp: we exhibit solutions for primes with $\text{leg}6p=+1$ realising each branch of the factorisation argument. All results are corroborated by extensive computational verification, and we state several open problems, including the complete classification of the solutions when $\text{leg}6p=+1$.