I will argue that the examples you gave are "simplest" in some strong sense, so although they look unnatural, if Martians study commutative algebra they will have to come up with them at some point.
Let's look at the first one $A=k[[x,y,z]]/(x^2-y^2, y^2-z^2, xy,yz,zx)$. Suppose you want
a $0$-dimensional Gorenstein ring which is not a complete intersection (complete intersections are the cheapest way to get Gorenstein but non-regular, for example $k[[x]]/(x^2)$).
Then it would look like $A=R/I$, where $R$ is regular and $I$ is of height equals to $\dim R$. If $\dim R=1$ or $2$, then $I$ would have to be a complete intersection, no good (now, the poor Martian may not know this at the begining, but after trying for so long she will have to give up and move on to higher dimension, or prove that result for herself). Thus $\dim R=3$ at least, and we may assume $R=k[[x,y,z]]$ (let's suppose everything are complete and contains a fied).
Since $I$ is not a complete intersection, it must have more than $3$ minimal generators. If it has $4$ then it would be an almost complete intersection, and by an amazing result by Kunz (see this answer), those are never Gorenstein.
So, in summary, a simplest $0$-dimensional Gorenstein but not complete intersection would have to be $k[[x,y,z]]/I$, where $I$ is generated by at least $5$ generators. At this point our Martian would just play with the simplest non-degenerate generators: quadrics, and got lucky!
(You can look at this from other point of view, Macaulay inverse system or Pfaffians of alternating matrices, see Bruns-Herzog, but because of the above reasons all the simplest examples would be more or less the same, up to some linear change of variables)
On to your second example, $k[[t^3,t^5,t^7]]$. Now very reasonably, our Martian wants
a one dimensional Cohen-Macaulay ring B that is not Gorenstein.
Since $\dim B=1$, being a domain would automatically make it Cohen-Macaulay. The most natural way to make one dimensional domains is to use monomial curves, so $B=k[[t^{a_1},...,t^{a_n}]]$. But if $n=2$ or $n=3$ and $a_1=2$ you will run into complete intersections, so $a_1$ would have to be $3$ and $n=3$ at least, and on our Martian goes...
(Again, you can arrive at this example by looking at things like non-symmetric numerical semi-groups, but again you would end up at the same simplest thing).
Reference: I would recommend this survey for a nice reading on Gorenstein rings.