There is a long exact sequence

$$0 \to H^1(G,Hom(M,N)) \to Ext_{\mathbb{Z}G}^1(M,N) \to \cdots $$
$$\begin{array}{lll}
\cdots & \to & H^i(G,Hom(M,N)) \to Ext_{\mathbb{Z}G}^i(M,N) \newline
& \to & H^{i-1}(G,Ext_{\mathbb{Z}}^1(M,N))\to H^{i+1}(G,Hom(M,N)) \to \cdots
\end{array}$$

For, as pointed out by Will and Mariano, there is a spectral sequence
$$H^i(G, Ext^j_\mathbb Z (M,N)) \Rightarrow Ext_{\mathbb ZG}^{i+j}(M,N)$$
and since the projective dimension of $\mathbb{Z}$ is one, the spectral sequence takes the form
$$E_2 = \quad
\begin{array}{ccccc}
\vdots & \vdots & \vdots & \vdots & \newline
0 & 0 & 0 & 0 & \cdots \newline
\bullet & \bullet & \bullet & \bullet & \cdots \newline
\bullet & \bullet & \bullet & \bullet & \cdots
\end{array}$$
Now the relation $E_\infty =E_3 = \ker(d_2)/\text{im}(d_2)$ yields the exact sequence
$$0 \to \ker(d_2^{i-2,1}) \to E_2^{i-2,1} \to E_2^{i,0} \to H^i \to \ker(d_2^{i-1,1}) \to 0$$
where $H^i=Ext_{\mathbb ZG}^i(M,N)$ is the abutment.

**Remark:** The statement remains true if we replace $\mathbb{Z}$ by an hereditary commutative ring $R$, $\mathbb{Z}G$ by an augmented $R$-projective $R$-algebra $A$ and $H^\ast(G,-)$ by $Ext_A^\ast(R,-)$.