In addition to the other answers, here is how the solutions can be derived using a graph representation.
Let A → B denote "the chest of metal A contains coins of metal B". Then we are looking to put seven arcs in a graph with seven nodes (one of each type of metal) such that each node has both an incoming and an outgoing arc.
From the last three clues, we have the following path:
silver → X → steel → Y → gold → Z → brass
At first glance, it would seem that all seven metals are represented here, in a single cycle linking all chests. But this is forbidden, since we know that the brass chest does not contain silver coins.
Thus we have to reuse metals, and the only way to resolve this is by setting:
X = brass and Z = silver
winding up with the five-metal cycle
silver → brass → steel → Y → gold → silver
Other attempts at instantiating variables will lead to coins of the same metal appearing in more than one chest (i.e. a node with more than one incoming arc in the graph representation).
This leaves the variable Y with three candidate metals: pewter, lead and tin. Here's where the first and third clues come in. If we set Y=tin, then there are two solutions:
the pewter chest can contain either pewter or lead coins, and the lead chest takes the remaining metal type.
Setting either Y=pewter or Y=lead we wind up with a single solution in each case:
the remaining two chests contain only coins of their own type.
Hence, we come to the four solutions given in the other answers.