Expressing everything in $s=x+y$ and $p=xy$, we get $p^2+s^2+1=12p$ and $p^2+2sp+s^2+2s+1=25p$, hence $2sp+2s=13p$. Solving for $s$, plugging in and simplifying we find that $4p^4-40p^3+81p^2-40p+4=0$. Now this is degree four, but symmetric, hence we can substitute $p+\frac{1}{p}$ to solve two quadratics. Even nicer, we can "guess" the factorization $(p^2-8p+4)(4p^2-8p+1)=0$. We then get $p=4 \pm \sqrt{3}$ and $p=1 \pm \frac{\sqrt{3}}{2}$. We can then solve for $s$ and then the resulting quadratic for $x,y$ to get the eight solutions $(2,2 \pm \sqrt{3}), (\frac{1}{2}, 2 \pm \sqrt{3})$ and the permutations.

In retrospect though, we have dismissed an additional symmetry in each of the variables, which leads to the following simpler solution: Write $u=x+\frac{1}{x}$ and $v=y+\frac{1}{y}$ to find $(u+2)(v+2)=27$ and $uv=10$, hence $u+v=\frac{13}{2}$ and hence $u,v=\frac{13 \pm 3}{4}$ so that we get $u=4, v=\frac{5}{2}$ or vice versa. From here, we just need to solve one more quadratic to get $x=2\pm \sqrt{3}$ and $y=2$ or $y=\frac{1}{2}$, or vice versa.