I want to find all the srg with parameters $(v,k,0,3)$. I leave my work so far below:
By the balloon equation we obtain that $v=\frac{k^2+2k}{3} + 1$.
Suppose that $k \geq \frac{v}{2}$. Then $k \in \{1,2,3\}$ but clearly we need that $k \geq b$ so $k$ must be $3$. This gives a graph with parameters $(6,3,0,3)$ which I know that exists.
Now suppose that $k < \frac{v}{2}$. The eigenvalue theorem say that $\sqrt{9 + 4(k-3)}$ must be an integer, but this still gives many possibilities. Eg. $k = 1,3,7,13,21,31 \ldots$ So I guess that the next step would be to consider the following equations:
$$ m_1 + m_2 = -k $$ $$ m_1 \lambda_1 + m_2 \lambda_2 = \frac{k^2+2k}{3} $$
where $\lambda_{1,2} = \frac{1}{2}(-3 \pm \sqrt{9 + 4(k-3)})$ and $m_1,m_2$ their respective multiplicities.
Well, the eigenvalues do not have to be integers. In that case, you have a $\textit{conference graph}$, so the multiplicities of $\lambda_1$ and $\lambda_2$ are equal. The only candidate that you obtain this way seems to be $k = 0$, so this will not lead to a strongly regular graph.
So assume now that the eigenvalues are integers, then there must exist some $z$ such that $z^2 = 9 + 4(k-3) = 4k-3$, as you noticed. We can now express $k$, $v$, $\lambda_1$ and $\lambda_2$ as polynomials in $z$.
Recall that the trace of a matrix equals the sum of its eigenvalues (weighted by multiplicity). The trace of an adjacency matrix is $0$, so we obtain that $m_1 \lambda_1 + m_2 \lambda_2 + k = 0$. Since $m_2 = v - 1 - m_1$, we can now obtain a long equation that will have the form $$z\cdot(\text{polynomial in $z$ and $m_1$ with integer coefficients}) = 27.$$ The key observation now is that both factors must be integers, so in particular $z$ must divide 27. That gives only a few candidates for $k$ for which you need to check whether such a strongly regular graph exists.