consider $\gcd(x,y) =d $ $x=da,y=db , \gcd(a,b) = 1$ $d^5a^5+d^4b^4 = p*d^2ab$ $d^3a^5+d^2b^4 = p*ab$ $d^2(b^4+da^5) = p*ab$ $a \mid d^2$ $b \mid d^2$ divide into two possibilities because i saw something wrong yesterday $ab \mid d^2$ $kab=d^2$ $k(b^4+da^5) = p$ implies $k = 1$ , $b^4+da^5 = p$ $ab=d^2$ implies b,a square since their gcd is 1 $a = m^2$ $b = n^2$ $d = mn$ $n^8+mn*m^{10} = p$ implies $n = 1$ $m^{11}+1 = p$ but by zsigmondy the equation has other sols if $m>1$ so the only sols is $m =1$ $a = 1,b=1,d=1$ $x = 1,y=1$ Here is the correction $a \mid d^2$, $d^2 \leq b \mid d^3$ $d^3 = ka*d^2t$ $b = d^2t$ $d = kat$ thus we have $b^4 +kat*a^5 = pa*t$ notice $\gcd(b^4+kat*a^5,a) = 1$ implies that $a = 1$ $b = k^2t^3$ $a = 1$ $d = kt$ $b^4+kt = pt$ $k^8t^12+kt = pt$ $k^8t^{11}+k=p$ implies$ k = 1, t^{11}+1 = p$ again use the same argument Zsigmondy thanks for the correction @below

@StarLex1: There's a flaw in your solution man. You have $b\mid d^3$ not $b\mid d^2$, which unfortunately makes your solution incomplete. Here's a complete solution. I clai $(x,y,p)=(1,1,2)$ is the only solution. Let $x=da$ and $y=db$ with $(a,b)=1$. We have $d^3 a^5 + d^2 b^4 = pab$. In particular, $a\mid d^2$ and $b\mid d^3$. Furthermore, $d^2\mid pab$. Now, we have by AM-GM inequality that $pxy= x^5+y^4\ge 2x^2y^2$, so $xy\le p/2$. Hence, $p\nmid ab$. If $p\mid d$, then, $p\mid d^2$, which in turn yields $p\mid ab$, a contradiction. So, $d^2\mid ab$. Now, let $d^2=ak$, so $k\mid b$. From here, we obtain that $(a,k)=1$. So, both $a$ and $k$ are squares, set $a=m^2$ and $k=n^2$ so that $d=mn$. Furthermore, $b=n^2\ell$ for some $\ell \mid d$. With these substitutions, we now obtain \[ m^2n^2(mn\cdot m^{10} + n^8 \ell^4) = pm^2 n^2\ell \Rightarrow m^{11}n + n^8\ell^4 = p\ell. \]Now, we have $\ell \mid n$. Furthermore, LHS is divisible by $n$, so $n\mid p\ell$. As $(n,p)=1$, we must have $n\mid \ell$, so that $n=\ell$. With these, we obtain $m^{11}+n^{11}=p$. If now $mn>1$ then $p>m+n\ge 2$ and $p$ is divisible by $m+n$, a contradiction. For $m=n=1$, we have $p=2$, and we recover $(x,y)=(1,1)$.

Nice problem! Answer: $(x, y, p)=(1, 1,2)$. All variables in future are natural numbers. Let $gcd(x, y)=d, x=ad, y=bd$, then $gcd(a, b)=1$. We have $d^2(a^5d+b^4)=pab$. If $p|d$, then $a^5d+b^4|ab$ and $ab \geqslant a^5d+b^4$, which impossible. So, since $p$ is prime, we must have $p|a^5d+b^4$, and so $d^2|ab$. Let $$a^5d+b^4=pk (1)$$, then $$ab=d^2k (2)$$. Not hard to see that if $d|ab$, then $d=gcd(a, d) \cdot gcd(b, d)$. Let $gcd(a, d)=v, gcd(b, d)=n, a=uv, b=mn$, so $d=vn$. Since $ab=d^2k$ we have $um=vnk$, so ($gcd(a, b)=1$) we have $v|u, n|m$. Let $u=vt, m=nl$, so $a=v^2t, b=n^2l$. Now let's see on (1): $v^{10}t^5vn+n^8l^4=pk$. From (2) we have $k=tl$, so $v^{10}t^5vn+n^8l^4=ptl$, or $n(v^{11}t^5+n^7l^4)=ptl$, so $t, l |LHS$. $gcd(tl, v^{11}t^5+n^7l^4)=1$, because $1=gcd(a, b)=gcd(v^2t, n^2l) (*)$, so $tl|n$ and $v^{11}t^5+n^7l^4|p$. But $v, t, n, l \geqslant 1$, so $v^{11}t^5+n^7l^4=p$ and $n=tl$. Also, we have $gcd(n, t)=1$ (look at (*)), so $t=1$ ($t|n$). Now, since $n=l, p=v^{11}t^5+n^7l^4$ we have $p=v^{11}+n^{11}=(v+n)\left(v^{10}-v^9n+v^8n^2-...+n^{10} \right)$. The second multiple is more than $1$ in all cases but $v=n=1$ ($v, n \in \mathbb(N)$) and $v+n>1$ always. So, the only solution is $p=2$, $v=n=1$. So, let's go back and achieve that $x=y=1, p=2$ the only solution.