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?

Here are the announced corrections to my first answer.
3) Beforehand we must come back to the original question : "For which values of $a\in \mathbf Q^*$ do integer solutions to the equation $x^2+x+1=a(y^2+1)$ (1) exist ?" As pointed out by @Gerry Myerson, this formulation does not make much sense (just take $a$ to be a quotient of the desired form), so I guess that the OP actually meant : "What can we say about the integer solutions of this equation (1) when they exist ?"
 Let us first study the rational solutions of (1) as in part 1). The solutions to the more general equation (3b) are parametrized by equation (3) , which is just a way to express that $a=N(z)$, where $N$ is the norm map of $\mathbf Q (\sqrt 3)(i)/\mathbf Q (\sqrt 3)$ and $z=(r+s\sqrt 3) + i(t+u\sqrt 3)$. If we fix $a$ and a solution $z_0$, all the solutions to $a=N(z)$ will be of the form $z=Z z_0$, with $N(Z)=1$ (this amounts to determine the pre-image of $a$). The solutions of the latter normic equation are in turn parametrized by $R^2+T^2+3(S^2+U^2)=1$  (4), which expresses the representation of $1$ by a $4$-variable quadratic form. There is a well established theory for that (local- global principle, Hilbert symbols, etc., see e.g. CF, op. cit.), but let us use our previous parametrization. To simplify the notations, write $z=[r,s,t,u]$ when $z=(r+s\sqrt 3) +i(t+u\sqrt 3)$, and analogously $Z=[R,S,T,U]$. Then $z=z_0Z =[r_0R-t_0T+3(s_0S-u_0U), r_0s_0+s_0R-u_0T-t_0U, t_0R+r_0T+3(u_0S+s_0U),t_0S+u_0R+r_0U+r_0U+s_0T]  (5)$. A not very enlightening parametrization, but note that there is obviously an infinite number of solutions (in (4), just take $U=S=0$ and solve $R^2+T^2=1$) 
4) Switching from the general equation (3) to the original equation (1), we  adapt slightly the argument of 3) to exploit the particular expression of $a= N_3(x_0-j)/N_1(y_0-i)$ (notations of (1))= $N((x_0-j)/(y_0-i)$. Put $z_0=(x_0-j)/(y_0-i)$; note that by identification one can take $(y_o^{2}+1)z_0 = [y_0(x_0 -\frac 12), \frac 12, x_0+\frac 12,- \frac {y_0}2]$ (6). To get all the rational solutions $(x,y)$, the approach in 3) works without any change. To catch the integral solutions, use (5) and (6), although the computations are not very instructive.
Remark on the Pell-like equation.  Putting $z=x+\frac 12$, equation (1) becomes a "Pell-like" equation $z^2-ay^2=a- \frac 34$ (7). But the difficulty is that $a$ appears on both sides; this is related to the OP formulation as discussed at the beginning of part 3). To go on, take $a$ as a parameter given by  a solution  $(z_0,y_0)$ of (7). There are then two possible approaches : 
(i) Algebraically, as in part 3), we are led to solve a normic equation $N(z-y\sqrt d)=N(z_0-y_0\sqrt d)$, where $d=a-\frac 34$ and $N$ is the norm map of the quadratic extension $\mathbf Q(\sqrt d)/\mathbf Q$. By Hilbert's 90 (or direct computation) any element of norm $1$ in such an extension can be written under the form $(e-f\sqrt d)/(e+f\sqrt d)$, and we are brougt back to parametrize $(x,y)$ from $(z-y\sqrt d)/(z_0-y_0\sqrt d)=(e-f\sqrt d)/(e+f\sqrt d)$ by identification. This seems rather neat, but it's actually not, because $a$ is hidden under two disguises, $d$ and $(z_0, y_0)$. Besides, the symmetry in calculations is broken by the eviction of the fields $\mathbf Q (i)$ and $\mathbf Q(j)$, which played such a prominent role in parts 1) and 2).
(ii) Geometrically, the homogeneous form of (7), say $aY^2 + dT^2 = Z^2$, is the equation of a projective conic $C$ defined over $\mathbf Q$ and admitting a rational point $M_0$ (coming from $(x_0, y_0)$). Algebraic geometry then tells us that $C$ is isomorphic to the projective line $\mathbf P^1(\mathbf Q)$. The isomorphism is given by drawing the line joining $M_0$ to some fixed well chosen point of the projective plane $P^2(\mathbf Q)$, and then taking its intersection with $ P^1(\mathbf Q)$ (embedded in $P^2(\mathbf Q)$ as a coordinate axis). This is exactly the same kind of parametrization as for, say, the pythagorean triples. However the final parametrization does not look neater than (i). 
