Note that $2=1^2+1^2$ and that a prime $p$ of the form $4n+1$ can be written as a sum of integer squares in essentially one way. eg $5=1^2+2^2, 13=2^2+3^2$. On your method of counting these become two ways.
Then note also that $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$$ by simple computation. This is related to $(a+ib)(c+id)=(ac-bd)+i(ad+bc)$ where $i^2=-1$, and that observation enables us to draw on deeper mathematical structure. But for the moment the simple observation is enough.
So, for example, we get $65=5\times 13=1+64=49+16$ by pairing the decomposition differently. That counts as four examples on your counting. Then introduce another factor (eg another $5$, or $17$) and you can continue to build examples where you have as many pairs of squares as you like.
Further examples come once you have a decomposition - you can multiply everything by any square number you choose..