Yes, we cheesed it.

python3 RsaCtfTool.py -n 0x3b7c97ceb5f01f8d2095578d561cad0f22bf0e9c94eb35a9c41028247a201a6db95f -e 0x10001 --uncipher 0x1B5358AD42B79E0471A9A8C84F5F8B947BA9CB996FA37B044F81E400F883A309B886 --private

Flag:

darkCTF{4v0iD_us1ngg_p1_pr1mes}

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As the name suggests, it's XOR.

We know the plaintext starts with darkCTF{ so using this as the key as so you'll get the actual key outputted- see here

Therefore using the key 1337hack gives us the flag- see here

Flag:

darkCTF{kud0s_h4xx0r}