On the Diophantine equation $Cx^2+p_1^{2m_1}p_2^{2m_2}\cdots p_k^{2m_k}=y^q$

Let $C>3$ be an odd square-free integer, $q$ an odd prime not dividing the class number of $\mathbb Q(\sqrt{-C})$, and $p_1,\dots,p_k$ distinct odd primes not dividing $C$. We describe all solutions of $Cx^2+p_1^{2m_1}\cdots p_k^{2m_k}=y^q$ in positive integers with $\gcd(x,y)=1$ and $y$ odd. Writing $N=p_1^{m_1}\cdots p_k^{m_k}$, every solution comes from a unique factorization $x\sqrt C+N\sqrt{-1}=\pm(u\sqrt C\pm v\sqrt{-1})^q$, or from a half-integral analogue when $q=3$, in which each $p_i\ne q$ divides exactly one of $v$ and $N/v$. Using primitive divisors of Lehmer sequences, we show that for $q\ge5$ some $p_i$ satisfies $p_i\equiv\leg{-C}{p_i}\pmod q$; in particular, there is no solution when $q\ge\max(5,p_1+2,\dots,p_k+2)$. For $q=3$ the equation reduces to four explicit auxiliary equations.