Can Ext over a group ring always be expressed as group cohomology ?  Given a group $G$ and $G$-modules $M,N$ with $M$ $\mathbb{Z}$-free then it's well known that 
$$Ext_{\mathbb{Z}G}^i(M,N) \cong H^i(G,Hom(M,N))$$
for all $i \ge 0$ (a reference is Brown, Cohomology of Groups, Proposition 2.2). 
But what happens if $M$ is not $\mathbb{Z}$-free ? Is it still possible to express $Ext_{\mathbb{Z}G}^i(M,N)$ by cohomology groups of $G$ (maybe in form of a spectral sequence) ?  
 A: 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 $Ext_A^\ast(R,-)$. 
A: As you've probalby noticed, we can't just take group cohoology. If $G$ is trivial, we are just taking exts in the category of abelian groups, which are nontrivial, but the group cohomology is trivial.
The spectral sequence idea is completely correct. The functor $Hom_{\mathbb ZG}(M,-)$ is the composition of the functor $Hom_{\mathbb Z}(M,-)$ with the functor $H^0(G,-)$. Here we need to see $Hom$ as a $G$-module. $G$ acts by taking $f$ to $g^{-1} \circ f \circ g$.
Now we use the spectral sequence for a composition of derived functors. $Hom_{\mathbb Z}$ preserves projectives so that makes sense. 
So we get a spectral sequence sending $H^i(G, Ext^j_\mathbb Z (M,N))$ to $Ext^i_{\mathbb ZG}(M,N)$.
