$k>1$ Let $gcd(x,y)=d$ then and $x=ad,y=bd, (a,b)=1$ $d^{k-1}a^kb=(a^2+b^2)p$ $a^2|d^{k-1}a^kb \to a^2|b^2p \to a^2|p \to a=1$ $d^{k-1}b=(1+b^2)p$ $b|p \to b=1,p$ $b=1 \to d^{k-1}=2p \to d=2,k=3$ and $(x,y,p)=(2,2,2)$ or $k=2,d=2p \to (x,y,p)=(2p,2p, p) $ $b=p \to d^{k-1}=1+p^2$ If $d=2$ $2^{k-1}=1+p^2$ $4\not | 1+p^2 \to k=2,p=1$ but $p$ is not prime If $d>2$ $d^{k-1}-1=p^2 \to d-1|p^2 \to d-1=p$ $p^2=(p+1)^{k-1}-1$ $p^2=(p+1)^{k-1}-1 >p^{k-1} \to k<3$ and for $k=2$ there are not solutions. So $k=2, 3$

RagvaloD wrote: and for $k=2$ there are not solutions. Are you sure? What about $x=y=2p$ with $p$ prime?

@above Thanks, I fixed

Suppose $k=1$. Then the given equation becomes $$xy=p(x^2+y^2)\ge p\cdot 2xy \iff 1\ge 2p,$$which is impossible. From now on assume $k\ge 2.$ Denote $d=\gcd(x,y)$, so $x=dx_1$ and $y=dy_1$, where $\gcd(x_1, y_1)=1$. Now the given condition transforms into $$d^kx_1^kdy_1=pd^2(x_1^2+y_1^2) \iff d^{k-1}x_1^ky_1=p(x_1^2+y_1^2).$$ Now we split the problem into 3 cases: Case $1: p\mid y_1$. Write $y_1=pm$ and hence $d^{k-1}x_1^km=x_1^2+y_1^2$. From here it follows that $x_1\mid y_1^2$. Suppose $x_1 > 1$. Let $q$ be a prime divisor of $x_1$. Hence, $q\mid x_1$ and $q\mid y_1^2$, so $q\mid y_1$, which is impossible since $\gcd(x_1, y_1)=1.$ Thus, $x_1=1$ and $x=d$, so $$d^{k-1}m=1+y_1^2=1+p^2m^2.$$Hence, $m\mid 1 \Rightarrow m=1$ and thus $y=dy_1=dp.$ Substituting these values for $x$ and $y$ in our original condition gives that $d^{k-1}-1=p^2$. Suppose $k=2$. Then $d=p^2+1,$ so $x=p^2+1$ and $y=p(p^2+1)$. A quick check gives that $(x,y)=(p^2+1, p(p^2+1))$ is indeed a solution for $k=2$. Now assume $k\ge 3$. Thus, $$d^{k-1}-1=(d-1)(d^{k-2}+d^{k-3}+\dots+d+1)=p^2$$and since the left bracket is smaller than the right bracket, we obtain that the only possibility is to have $d-1=1,$ i.e. $d=2$ and $d^{k-2}+d^{k-3}+\dots+d+1=p^2$. Hence, we get $2^{k-1}-1=p^2$, however $$p^2 = 2^{k-1}-1\equiv -1 \equiv 3 \pmod{4},$$which is impossible. Finally, $k=2$ is a solution. Case $2: p\mid x_1$. Write $x_1=pn$ and hence $$d^{k-1}p^kn^ky_1=p(p^2n^2+y_1^2) \iff d^{k-1}p^{k-1}n^ky_1=p^2n^2+y_1^2.$$Since $k\ge 2$, we must have that $p\mid y_1^2 \iff p\mid y_1$, which is impossible because of $\gcd(x_1,y_1)=1.$ Case $3: p\mid d^{k-1}$. Therefore, $p\mid d$. Write $d=pd_1$, so the given condition is equivalent to $$p^{k-2}d_1^{k-1}x_1^ky_1=x_1^2+y_1^2.$$From here we must have $y_1\mid x_1^2$. Assume $y_1>1$ and let $q$ be a prime divisor of $y_1$. Hence, $q\mid x_1$, so again we reach to a contradiction since $\gcd(x_1,y_1)=1.$ Thus, $y_1=1$. From here $x_1\mid x_1^2+1$, so $x_1=1$. Now we obtain $$p^{k-2}d_1^{k-1}=2.$$Now we will only consider $k\ge 3$ (we have already shown that $k=2$ is a solution). Suppose $k=3$. Then we must have $pd_1^2=2 \Rightarrow p=2$ and $d_1=1$. Hence, $d=2$ and $x=y=2$. A quick check shows that $(x,y)=(2,2)$ is indeed a solution. Now if $k\ge4$, we get and immediate size contradiction, so finally the only answer in this case is $k=3$. Finally, the solution is $k\in \{2,3\}$.