One may modify Richard's attempt to get an example, using a method of Stallings.
Consider the Baumslag-Solitar group $BS(1,2)=\langle a,b | bab^{-1}=a^2\rangle$, with homomorphism $\phi: BS(1,2) \to H_1(BS(1,2)) = \mathbb{Z}$ given by $b\mapsto 1, a\mapsto 0$. Then $ker(\phi)=\langle b^k a b^{-k} \rangle \cong \mathbb{Z}[\frac12]$, so $ker(\phi)$ is infinitely generated. Take the double of $BS(1,2)$ along $ker(\phi)$, $Dub(BS(1,2),ker(\phi))$. This gives a group with infinite presentation $$Dub(BS(1,2),ker(\phi))=\langle b_1,b_2,a | b_1ab_1^{-1}=a^2=b_2ab_2^{-1}, b_1^kab_1^{-k}=b_2^k a b_2^{-k}\rangle.$$ This group has $H_2(Dub(BS(1,2),ker(\phi)),\mathbb{Z})$ infinitely generated, because of the Mayer-Vietoris sequence $$H_2(Dub(BS(1,2),ker(\phi)) \to H_1(ker(\phi)) \to H_1(BS(1,2))\oplus H_1(BS(1,2)).$$
Now, there is an automorphism of $BS(1,2)$ fixing $b\mapsto b$ and sending $a\mapsto a^2$ (corresponding to multiplication by $2$ on the subgroup $ker(\phi)\cong \mathbb{Z}[\frac12]$). This automorphism extends to $Dub(BS(1,2), ker(\phi))$. Take the mapping torus of this automorphism gives the group presentation $$\langle b_1,b_2,a,x| b_1ab_1^{-1}=a^2=b_2ab_2^{-1}, xb_1=b_1x, xb_2=b_2x, xax^{-1}=a^2\rangle.$$ This group then is finitely presented, whose kernel of homomorphism to $\mathbb{Z}$ given by $x\mapsto 1, a, b_1, b_2\mapsto 0$ is finitely generated with infinitely generated $H_2(Dub(BS(1,2),ker(\phi)), \mathbb{Z})$.