If you care about all solutions and not just some nice ones, especially if you are using a computer to do the grunt work, fire up any CAS (computer algebra system) at your disposal. Any self-respecting CAS has an implementation of groebner basis algorithms (to use for problems like this under the hood!) - but try sympy! The lex order groebner basis of your system is
$$52704x - 6250y^5 + 27700y^4 - 90775y^3 + 220495y^2 - 297625y + 304567=0$$
$$10y^6 - 30y^5 + 123y^4 - 246y^3 + 408y^2 - 480y + 215=0$$
The last equation is a univariate six-degree polynomial that factors into
$$(y-1)(y+i\sqrt5)(y-i\sqrt5)(y-1/2 + 9i/10\sqrt5)(y-1/2 - 9i/10\sqrt5)$$
So like your solution we have $y=1$ as well. Backsubtition into the first equation (of the groebner basis) gives $$52704x + 158112=0$$ but $-158112/52704=-3$. So indeed $y=1$ and $x=-3$ is a solution...
This fits nicely with what Dietrich posted: As he points out if $x=2$ then $y$ is a solution to $y^2=-5$ both of these solutions (the positive and negative solution of that quadratic) also solve the equation in the groebner basis that only contains $y$ (and not $x$). Which you also see from the factorization. Substituting $y=i\sqrt5$ into the first equation gives $$52704x - 105408=0$$ and we have $105408/52704=2$ so we have another nice solution of $y=i\sqrt5$ and $x=2$! The same for $y=-i\sqrt5$.