If you move things around and square the equations, you have the following two starting equations.
If you plug one of these into the other and factor out a term, you get
This is a quartic equation, with four solutions in general. There is a linear term, $x-9$, and the remaining cubic term. If $x=9$, the linear term goes to zero and the equation is solved. If you plug $x=9$ back into the original equations, you get $y=4$, so this represents the original solution you found. To find the other solutions, we only have to focus on the roots of the remaining cubic term. So we're left with finding the roots of this equation
All cubics have three roots, if you count complex roots and double roots. The possibilities go like this for any cubic:
There are three real, distinct roots
There are three real roots, but two of them are merged into a double root
There is one real root and a pair of complex conjugate roots
You can follow along with this procedure from Wikipedia to calculate the exact roots of the equation. Or you can use a computer solver for the roots that does a similar procedure internally:
This shows you that this cubic is in the first family with three real, distinct roots, with $x \approx 7.8687, 12.848, 14.283$ and corresponding $y\approx 9.80504, 3.4151, 10.7781$.
As a user pointed out, the values we just found should be plugged into the original equations to see if they actually solve the equation, or if they are extraneous. It turns out that these three solutions are indeed extraneous, meaning that they are spurious results of squaring the original equations. This leaves the only solution we've found as $(9,4)$. It is necessary that the solutions are a subset of the four presented, because the system is fundamentally fourth order (two quadratic equations), so it can have at most four solutions. The only one of the necessary solutions that is sufficient is $(9,4)$, so this is the unique solution. It consists of two integers, so we have proof that this equation only has integer solutions.