Let $r_2(n)$ be the number of representations of $n$ as a sum of two squares, and let $l(n)$ be the number of ways to write $n$ as $x^2+xy+y^2$. Then as you mentioned we have that $$\sum_{y,x\in \mathbb{Z}} q^{x^2+y^2}=\sum_{n=0}^\infty r_2(n) q^n \ \text{and} \ \sum_{y,x\in \mathbb{Z}} q^{x^2+xy+y^2}=\sum_{n=0}^\infty l(n) q^n.$$
Exact number of representations: We can evaluate explicitely write down $r_2(n)$ and r_2(n)$, the number of representations of $l(n)$ explicitly using some ideas already mentioned. (We need n$ as a Theorem sum of Fermat and a theorem two squares. See Sum of Jacobi regarding which primes can be represented.)
However Squares Function.
Also, in a similar manner we can explicitely write down $l(n)$, the number of representations of $n$ as $x^2+xy+y^2$. See the answer on this math stack exchange post.
Theta Functions: We can evaluate the infinite series in terms of Jacobi theta functions without much difficulty. To deal with the sum of squares, notice that
$$\sum_{y,x\in \mathbb{Z}} q^{x^2+y^2} =\left( \sum_{n=-\infty}^\infty q^{n^2}\right)^2= \vartheta_3(q)^2.$$ Next, we can transform $x^2 +xy+y^2$ into $\frac{n^2+3m^2}{4}$ where $m,n$ must have the same parity. Then $$\sum_{x,y}q^{x^{2}+xy+y^{2}}=\sum_{\begin{array}{c} x,y\ both\ odd \end{array}}q^{\frac{x^{2}+3y^{2}}{4}}+\sum_{\begin{array}{c} x,y\ both\ even \end{array}}q^{\frac{x^{2}+3y^{2}}{4}} $$
$$=q\sum_{n,m\in\mathbb{Z}}q^{n(n+1)+3m(m+1)}+\sum_{n,m\in\mathbb{Z}}q^{n^{2}+3m^{2}}.$$This can be rewritten in terms of the jacobi theta functions as $$\vartheta_2(q)\vartheta_2(q^3)+\vartheta_3(q)\vartheta_3(q^3).$$
Hope that helps,
