Let $$s(n) = \#\{(x,y) \in \mathbb{Z}^2 : x^2 + 3y^2\ = n\}$$ for $n \geq 1$ an integer and let $K = \mathbb{Q}(\sqrt{-3})$. The class number of $K$ is $1$ and the ring of integers $\mathcal{O}_K$ contains $2$ units, so it should be that $$s(n) = 2 \cdot \#\{I \subset \mathcal{O}_K \textrm{ ideal} : N(I) = n\},$$ where $N$ denotes the ideal norm. Thus we have $$\sum_{n=1}^{\infty}\frac{s(n)}{n^s} = 2\sum_{I \subseteq \mathcal{O}_K}\frac{1}{N(I)^s},$$ where $I$ runs over all ideals of $\mathcal{O}_K$. Since the ideal norm is completely multiplicative, we have an Euler product $$\sum_{I \subseteq \mathcal{O}_K}\frac{1}{N(I)^s} = \prod_{P \subseteq \mathcal{O}_K}\frac{1}{1 - N(P)^{-s}},$$ where $P$ runs over all prime ideals of $\mathcal{O}_K$. Each prime ideal of $\mathcal{O}_K$ lies above some integer prime $p$. Since the discriminant of $K$ is equal to $-3$ which is congruent to $5$ modulo $8$, the prime $2$ is inert in $\mathcal{O}_K$. So if $p \equiv 2\pmod 3$ is prime, then $p$ is inert in $\mathcal{O}_K$, so there is a unique ideal of $\mathcal{O}_K$ above $p$ and it has norm $p^2$. If $p \equiv 1\pmod 3$ is prime, then $p$ splits in $\mathcal{O}_K$, hence there are two ideals above $p$ and both have norm $p$. The prime $3$ ramifies in $K$, thus there is a unique ideal above $3$ and it has norm $3$.
Let $\chi$ be the odd Dirichlet character modulo $3$, i.e. $\chi(n) = 1$ if $n \equiv 1 \pmod 3$ and $\chi(n) = -1$ if $n \equiv 2 \pmod 3$. For a prime $p$ we have $$(1 - p^{-s})(1 - \chi(p)p^{-s}) = \begin{cases} (1 - p^{-s})^2 & \text{if } p \equiv 1 \pmod 3 \\ (1 - p^{-2s}, & \text{if } p \equiv 2 \pmod 3 \\ (1 - p^{-s}) & p = 3 \end{cases}$$ Therefore we should have $$\prod_{P \subseteq \mathcal{O}_K}\frac{1}{1 - N(P)^{-s}} = \prod_{p}\frac{1}{(1 - p^{-s})(1 - \chi(p)p^{-s})} = \zeta(s)L(s,\chi),$$ where $\zeta(s)$ is the Riemann zeta function and $L(s,\chi)$ is the Dirichlet $L$-function of $\chi$. But the rihgt-hand side as an $L$-series is $$\zeta(s)L(s,\chi) = \sum_{n=1}^{\infty}\left(\sum_{d \mid n}\chi(d)\right)n^{-s}.$$ Thus $$\sum_{n=1}^{\infty}\frac{s(n)}{n^s} = 2\sum_{n=1}^{\infty}\left(\sum_{d \mid n}\chi(d)\right)n^{-s},$$ so we have $$s(n) = 2\sum_{d\mid n}\chi(d).$$
But this is just false. We check that there are $6$ integer solutions to $x^2 + 3y^2 = 4$, namely $(x,y) = (\pm 2,0),(\pm 1,\pm 1)$. On the other hand, we have $$2\sum_{d \mid 4}\chi(d) = 2(\chi(1) + \chi(2) + \chi(4)) = 2.$$ I cannot for the life of me figure out what is wrong with the above argument, but obviously something is wrong because the resulting statement is just not true.
I realized the problem was in the first step since $-3 \equiv 1 \pmod 4$.