If $p=q=2$, obviously $r=2$ and $n=1$ because r must be prime. If $p,q\ne 2$ then $r$ must be even. That means $r=2$ and obviously no solution. If $q=2, p\ne 2$ we divide this into 2 cases. Case 1 $n$ is even, Let $n=2k$ There are exists $u,v\in\mathbb{N}$ such that $p^k=u^2-v^2, 2^k=2uv, r=u^2+v^2$ But $uv=2^{k-1}$ So $v=1, u$ is power of 2. Let $u=2^t$ So $p^k=2^{2t}-1=(2^t+1)(2^t-1)$ But $(2^t+1,2^t-1)=1$ so $t=1, p=3, k=1$ Substitute these in the initial equation we get a contradiction. Case 2 $n$ is odd and $n>3$ We can easily prove that $p^n+2^n>p^3+8>(p+2)^2$ But $p+2\mid p^n+2^n$ So $r\mid p+2$ But $r^2=p^n+q^n>(p+2)^2\ge r^2$. which is obviously impossible so $n=1$.