$M \otimes_Z N =0$ if and only if $M \otimes_R N =0$? If $R$ is a commutative ring, and $M$ and $N$ are $R$-modules, is $M \otimes_Z N =0$ if and only if $M \otimes_R N =0$?
The forward direction seems clear.  The backward direction seems like it should be false, but it also seems like it should follow from the following:
$(M \otimes_R N) \otimes_Z R = M \otimes_R (N \otimes_Z R) = (N \otimes_Z R) \otimes_R M = N \otimes_Z (R \otimes_R M)
= N \otimes_Z M = M\otimes_Z N.$
And is there a way to see this is true, or false, using the categorical definition of $\otimes_R$?  If $M \otimes_R N$ is thought of as a coequalizer, is there any reason why $M\otimes_R N=0$ would imply $M\otimes_Z N$?
$M\otimes_Z R \otimes_Z N \rightrightarrows M \otimes_Z N \rightarrow M \otimes_R N.$
 A: Here's a counterexample. Let $R = k[x]$ for $k$ a field. Then 
$$k[x]/(x - a) \otimes_{k[x]} k[x]/(x - b) \cong 0$$
for $a \neq b$ because the supports of the two modules are disjoint, but 
$$k[x]/(x - a) \otimes_{\mathbb{Z}} k[x]/(x - b) \cong k \otimes_{\mathbb{Z}} k$$ 
can be quite large, e.g. if $k = \mathbb{Q}$ then it is $\mathbb{Q}$. 
The mistake in your argument is being hidden by the fact that your notation doesn't make module structures explicit: if $N$ is an $R$-module, then $N \otimes_{\mathbb{Z}} R$ has not one but two different $R$-module structures (one coming from acting on $N$ and one coming from acting on $R$) and you switch between them in the course of the argument. Depending on where you think $N \otimes_{\mathbb{Z}} R$ has which $R$-module structure the argument fails at either the second or third "equality" (you should be talking about natural isomorphisms anyway). 
This is one of the perils of learning about tensor products of modules over commutative rings before learning about tensor products of modules over noncommutative rings. Over noncommutative rings it's harder to make mistakes like this because by default you lose more structure when you tensor and it becomes more important to make bimodule structures explicit and distinguish left and right actions. (Exercise: write down the correct statement of the associativity of the tensor product for bimodules over noncommutative rings.) 
