It's a bit too long for a comment: Brady's example is not coherent. Below is a sketch of the proof. The example constructed by Brady (see the reference in HJRW's answer) is obtained as a certain ramified covering over the space $X=\Theta^3$, where $\Theta$ is the theta-graph (two vertices and three edges). This graph is the union of two circles $C_1, C_2$. In particular, the space $X$ contains two 3-dimensional tori $C_1^3, C_2^3$, whose intersection contains a 2-dimensional torus $T$. Taking a ramified covering $Y\to X$, we obtain lifts $M_1, M_2$ of the tori $C_1^3, C_2^3$ to $Y$, such that $M_1, M_2$ are 3-dimensional compact hyperbolic manifolds intersecting in a subset containing a compact hyperbolic surface $S$ which is incompressible in both $M_1, M_2$ (the surface $S$ is a lift of the torus $T$). Furthermore, the manifolds $M_1, M_2$ are $\pi_1$-injective in $Y$. In particular, $\pi_1(Y)$ contains the amalgam $\pi_1(M_1)\star_H \pi_1(M_2)$, where $H$ is a certain quasiconvex subgroup of infinite index in both $\pi_1(M_1), \pi_1(M_2)$. Now, one uses the fact that both manifolds $M_1, M_2$ fiber over the circle, let $G_1, G_2$ denote surface fiber subgroups in $\pi_1(M_1), \pi_1(M_2)$. The intersection $G_1\cap H=G_2\cap H=F$ is a free group of infinite rank. Thus, the subgroup
$$
G=\langle G_1, G_2\rangle= G_1\star_F G_2
$$
is finitely generated but not finitely presentable (this can be checked, for instance, by looking at the 2-nd cohomology group of $G$ using the Mayer-Vietoris sequence).

I omitted many details in this proof but they are not hard to check once one understands details of Brady's construction.