$p^2 + pq+q^2=n^2$ [duplicate]

Since $p$ and $q$ are primes, $(pq)^{1/4}$ is not an integer. The fundamental theorem of arithmetic (or unique prime factorization theorem) immediately implies taht the only factorizations of $pq$ in positive integers are $p\cdot q$ and $1\cdot(pq)$, so the argument is correct.

All solutions to $$ p^2 + p q + q^2 = k^2 $$ with $p,q,k > 0$ and $\gcd(p,q,k) = 1$ but no requirement about primes, are given by $$ p = u^2 + 2 u v, \; \; q = v^2 - u^2, \; \; k = u^2 + u v + v^2, $$ with coprime $v > u > 0.$ See 120 DEGREES.

Since $p = u(u+2v), \; \; q = (v-u)(v+u),$ the only way to get both prime is $u=1, v=2,$ because we need both $u=1, \; \; v-u=1.$