Class-number obstruction from a ramified odd exponent ($x^2+d=y^p,p\mid d$)

Let $d$ be a positive square-free integer congruent to two modulo four and $p\ge 3$ a prime. If $p$ does not divide $h(\mathbb{Q}(\sqrt{-d}))$ and $p\mid d$, then $x^2+d=y^p$ has no positive integer solution. More generally, if $\gcd (x,a)=1$ and $x^2+a^{2}d=y^p$, the two conjugate principal ideals are coprime. Parity excludes a shared prime above two; for an odd common prime, divisibility of both conjugates would force the underlying rational prime to divide both $x$ and $d$, giving valuation one on the left side, incompatible with a $p$th power.

Unique prime ideal factorization consequently gives $(x+a\sqrt{-d})=A^p$. The class-number condition forces $A$ principal; the only units are $\pm 1$, both odd powers. Hence $x+a\sqrt{-d}=(u+v\sqrt{-d}{)}^p$. When $a=1$ and $p\mid d$, the coefficient of $\sqrt{-d}$ on the right is divisible by $p$: its linear term has coefficient $p$, and every higher odd term contains $d$. It cannot equal one. Conversely, an actual solution under these conditions forces divisibility of the class number by $p$. For example $13^2+174=7^3$ proves $3\mid h(\mathbb{Q}(\sqrt{-174}))$.