If it is a continuous LP then the optimizer should give you an infeasibility certificate, ie. a (small, if you're lucky) set of conflicting constraints. Finding a certificate is not any different than optimizing and is typically just one of the conclusions the optimizer can reach. Check for instance "Farkas lemma". From your description ("past a certain size") it could also be numerical issues or the like that come into play, especially if you actually expect it to always be feasible.

$t\geq x^{-m}$ is $t\geq z^m$ combined with $zx\geq 1$.

No, $t\geq xy$ is not convex, yet alone conic. That's why the slightly tricky and at first nonintuitive approach to fit it into the right form.

Note that the formulation of $t\geq x^3$ will force/require $x$ to be nonnegative.

Every respectable modeling tool (CVXPY, CVX, Yalmip) will perform this reformulation for you behind the scenes.

$t\geq x^2$ is SOCP, and therefore also $t\geq x^4$ by composing two of those ($t\geq y^2, y\geq x^2$) and $t\geq x^3$ is equivalent to $tx\geq y^2, y\geq x^2$, both of which are SOCP.

My back-of the-envelope calculations show that $n=143$ is the first counterexample to the contractibility of the complex $X_I$ from Andy's answer. $X_I$ is the clique complex of the graph where $ij$ is an edge when $gcd(i,j)>1$. When $N[u]\subseteq N[v]$ then $u$ can be removed from the clique complex without changing homotopy (standard flag complex stuff). A bit of coding then shows this sequence of folds leads to the cross-polytope on $\{2\cdot 3\cdot 7, 2\cdot 3\cdot 11, 2\cdot 3\cdot 13, 7\cdot 11, 7\cdot 13, 11\cdot 13\}$, generating $H_2$. I can write up a full answer later if necessary.

Your argument doesn't address the fact that the other cliques (except for the initial one) can intersect. That causes a problem in the "proceeding so on" stage.