Suppose $x,y\not\in \{1\}$ (otherwise is fairly straightforwards). Now we must have at least one of $x,y\mid x+y$ (else we will have a a composite product). WLOG $y=kx$. So $\tfrac{k^4x^3}{k+1}=p$ It is now again easy since $k+1<k^2$ for all but a few values of $k$, which are bashable.

Pretty sure only solution is $(14,2,7)$. Are there any other?

let (x,y) equal to z then x is equal to r*z and y equal to s*z where (r,s) equals 1 then we can show that s equals 1 and p equals y^3 - 1 which can be factored into (y-1 )*(y^2 plus y plus 1) unless y equals 2 .so only one solution (14,2,7) PS:SRY cannot use latex