Why can't we find $d$ if we know $e$ and $n$ in RSA? Say, we have RSA public key $(n,e)=(221,7)$,

and the encryption of the message $m=12$ is $$c=m^e \bmod n= 12^7 \bmod (221) = 194 \bmod (221)$$  

Can we find the $d$ by solving the equation;

$$194^d \bmod(221)=12$$  

I get that the numbers I used are way smaller than they are in real life, but this should be easier then just guessing $p$ and $q$ from $n$ right?
There are "hidden unknowns" in the notation. Suppose that $n$ is the modulus, $e$ the public exponent, $d$ the private exponent. Given a message $m$, its encrypted value is $c = m^e \bmod n$ (I am ignoring padding here), and you can certainly choose a message $m$ and compute $c$ given the public elements $e$ and $n$ (that's how RSA encryption works). Now, you might want to find $d$ by solving the following equation:
$$ c^d = m \bmod n $$
with only $d$ being unknown.

When working with real numbers, an equation of the form:
$$ a^x = b $$
for given values $a$ and $b$, is solved as:
$$ x = \frac{\ln b}{\ln b} $$
and you could imagine that "solving RSA" is just a matter of computing logarithm with the required precision.

However, this would neglect an important part, which is somewhat hidden in the notation. When we work "modulo $n$", we are actually having the following:
$$ c^d = m + kn $$
for the known ciphertext $c$, message $m$ and modulus $n$, unknown private exponent $d$, and unknown integer $k$. At that point, you have a single equation with two unknowns, and the usual tools of analysis with real numbers (e.g. logarithms) won't help at all.

In all generality, RSA is expressed in algebraic things such as rings of integers modulo $n$, and all equation solving activities in such things have their own rules, quite distinct from what you may have learned with real numbers. Obtaining $d$ from $e$ can be done relatively easily when $n$ is prime, but, of course, the crux of RSA is that $n$ is not prime.

We do not have any proof that factoring $n$ into its factors is really required to recover an unknown plaintext from the ciphertext. We do know that recovering the private exponent $d$ is equivalent to factoring $n$ (i.e. if you factor $n$ you can easily obtain $d$ from $e$, and if you somehow obtained $d$ and $e$ you can easily factor $n$); but there could be a method by which a ciphertext can be decrypted without revealing the private exponent itself. Right now, factoring $n$ is the best attack we know of; and factoring big integers is a problem that has been studied for quite some time (more than 2500 years, in fact), and the only sure thing we know is that there is no obviously easy solution. There might be an efficient factoring method, but it is definitely not obvious.