I found the generalization that $10^{2n} \equiv 2^n \pmod{49}$, but I haven’t been able to prove it or guarantee that it holds for all nonnegative integers. Is there a reason this holds?
Another approach got me $10^n \equiv 6 \pmod 7$ for $n = 6k + 3$, but then I got stuck (beyond just trying all valid n to see which $10^n \equiv 34 \pmod{49}$)
Wolframalpha shows that $x = 42n + 33$ but I would like an more elegant solution (if there is?)
Additionally, in general, how would one be able to determine how many values of $10^n$ it takes to complete one mod $x$ cycle? I know $10^n \pmod 7$ repeats every $6$ from fermat and mod $49$ repeats every $42$ (from wolfram alpha…), but is it just a coincidence that $6*7=42$?
Welcome to MSE! As I mentioned in my comment, finding the multiplicative order of $a \pmod m$ is a problem with no simple solution or formula. Similarly, finding the discrete logarithm (base $r$) of $a \pmod m$, which is:
$$x : r^x \equiv a \pmod m$$
has no simple solution. There are good algorithms that run rapidly if $\varphi(n)$ has certain properties (i.e., being smooth to a small prime), but no formula one can write.
The good news here is that once you've found the multiplicative order and the discrete log, you can find your answer using them, in exactly the form you got from WA. $42$ is the MO of $10 \pmod{49}$, and $33$ is the discrete log to base $10$ of $34 \pmod{49}$. Hence the family of solutions is $x \equiv 33 \pmod{42} \ =42n+33, n \in \mathbb{Z}$; or, to generalize, the solution of $a^x \equiv b \pmod m$ is:
$$x \equiv \log_{\text{ord}_m(a)} b \pmod {\text{ord}_m(a)}$$