Exponential cipher theory misunderstanding Say I pick modulus $m=2633$ and key $e=13$

My private key $d$ is $13$ inverse in $\bmod 2632$ which I need to compute

$(e,m)$ is my public key that anyone can use to encrypt message for me and then send it to me as:

encrypt message: "TH" ($T=19, H=07$) as $1907^{13} \bmod 2633$

but since $(e,m)$ is public anyone can compute my private $d$ using EEA

so where is the security? I know I do not understand this, could anyone explain?
Exponential cipher is a symmetric-key algorithm, not public-key one. So you don't have any public and private keys. You only have one secret key and it's used for encryption of messages between people that know the key.

The security of exponential cipher is not in the computiation of $d$ from $e$, which is easily computable by EEA as you said. The security of this cipher lies at the principle of discrete logarithm problem, as well as Diffie-Hellman key exchange does.

The principle:

You have prime modulus $m = 2633$, secret key $e = 29$, where $\gcd(e,\ m-1) = 1$ so $e$ has inverse $\textrm{mod}\ m-1$ and you can compute your key $d$ for deciphering:

$d = |e^{-1}|_{m-1} = |29^{-1}|_{2632} = 2269$

The only thing you have to keep secret is the key $e$.

The encryption pattern is ($c$ for ciphertext block, $p$ for plaintext block) :

$c = |p^e|_m$.

So you can encrypt "TH" (T=19, H=07) as $c = |1907^{29}|_{2633} = 2199$.

To decrypt:

$p = |c^d|_m = |2199^{2269}|_{2633} = 1907$ = "TH"

Discrete logarithm problem:

Consider a possible attacker who has modulus $m = 2633$, block of ciphertext $c = 2199$ and corresponding block of plaintext $p = 1907$. What does he know:

$c = |p^e|_m$

$2199 = |1907^e|_{2633}$

which means $e =|\log_{1907}{2199}|_{2633}$

The discrete logarithm problem says that it's not possible to compute the key $e$ from this equality in reasonable time (for higher numbers of course).