For which values of $a\in\mathbb{Q}$ does integer solutions to $x^2+x+1=a(y^2+1)$ exist? I am unable to determine for which values of $a\in\mathbb{Q}$ does integer solutions to
$$x^2+x+1=a(y^2+1)$$
in the form $(x,y)$ exist.
My initial idea was to set $a=\frac{c}{d}$ to get $dx^2+dx+d=c(y^2+1)$ for $c,d\in\mathbb{Z}$ but I am unsure how to convert this into Pell's equation nor do I know how to apply initial solutions. Any suggestions?

Surely a zest of Galois theory could do no harm. 
1) Consider first the equation with rational parameter and variables : $x^2+x+1=a(y^2 +1), a, x, y \in \mathbf Q^*$  (1). With $i^2=-1$ and $j^3=1$, (1) is equivalent to $a= N_3(x-j).N_1(y-i)^{-1}$, where $N_n$ denotes the norm map of $\mathbf Q(\sqrt -n)/\mathbf Q$. The natural question is then how to characterize the elements of $\mathbf Q^*$ which are products of two norms as above. This is a purely Galois theoretic exercise (see e.g. Cassels-Fröhlich, ex. 4.4, p. 358)  : Let $K$ be any field of characteristic $\neq 2$. An element of $a \in K$ is a product of a norm from $K (\sqrt b)$ and a norm from $K(\sqrt c)$ iff $a$, as an element of $K(\sqrt bc)$, is a norm from $L=K(\sqrt b , \sqrt c)$  (2). In our case where $\mathbf Q(\sqrt -1)=\mathbf Q(i)$ and $\mathbf Q(\sqrt -3)=\mathbf Q(j)$, (2) is equivalent to "$a \in \mathbf Q (\sqrt 3)$ is a norm from $\mathbf Q (\sqrt 3)(i)$", or "$a$ is a sum of two squares in $\mathbf Q(\sqrt 3)$", say $a=(r+s\sqrt 3)^2 + (t+u\sqrt 3)^2$. Since $a \in \mathbf Q$, a straightforward calculation shows that the latter condition is equivalent to $a$ is of the form $r^2+t^2+3(s^2+u^2)$, with $r, s, t, u \in \mathbf Q$ such that $rs=-tu$  (3). Summarizing : (3) is equivalent to  (3b)  $a= (m^2+mn+n^2)/(\mu^2 +\nu^2)$, with rational parameter and variables. Cp. the edit of @Tito Pezias III.
2) Let us apply this approach to the special expression of $a$ given by (1). Elementary calculations show that the relation $rs=-tu$ in (3) boils down to $(2x+1)y=0$ and we have two different cases which correspond each to a single normic equation :  (i) $y=0$, $a= x^2+x+1= N_{3} (x-j)$ ; (ii) $x=-\frac 1 2$,  $3 a^{-1}=4(y^2 +1)= N_{1}(2(y-i))$. 
Let us finally come back to the original question, with $x, y \in \mathbf Z$. In case (i), $a$ is a norm from  $\mathbf Z[j]$, but this is only a necessary condition. In case (ii), $3 a^{-1}$ is a norm from $\mathbf Z[i]$, i.e. $3 a^{-1}$ is a sum of two squares of integers, but again this is only a necessary condition; actually the inequality $3/4a\ge1$ cannot hold if $a\in \mathbf Z$. In the end, the integrality condition imposed on the variables $x, y$ seems rather irrelevant. A more natural condition could be that $x, y\in \mathbf Z$ and $a\in\mathbf Q^*$ is defined up to a square, see (3b).
Edit : Because of the OP's uncertainty, I checked again the "elementary calculations" in part 2) of my answer... and I found a sign error which completely destroys the conclusion, so there is no reduction to two normic equations in one variable. I replace 2) by 3) and 4) below, following the same method but getting a different final result. I put in another answer because it seems that typo difficulties appear when the post is too long. 
