Are ciphertext-only attacks on LFSRs possible?
For text known to be ASCII encoded as octets with high-order bit of octets at zero, that reveals one bit of the output of the LFSR out of 8. It allows finding the original state of the LFSR from (say) the first 51 octets of the ciphertext by merely solving a system of linear boolean equations; then decipher the rest. There is no guesswork involved.

For 7-bit plus parity (where the high-order bit of each octet is the eXlusive-OR of the 7 others, or the complement of that), linear algebra works too.

Things get more difficult as the plaintext becomes less redundant. With 7-bit ASCII, noticing that space, digits, common punctuation and all lowercase letters are in range $[\mathtt{0x20}\dots\mathtt{0x3F}]\cup[\mathtt{0x60}\dots\mathtt{0x7F}]$, we know that every such 7-bit symbol has bit with weight 5 set, and there is fair chance that 51 such characters occur in a row somewhere in the 800-character plaintext. A guess of where allows to decode the rest, and it is easy to detect if the guess was right or not, thanks to characteristics of English text.

More generally, if 51 bits at almost any position can be guessed, then the rest is uniquely determined with basic linear algebra, and the guess is verifiable by checking that the resulting plaintext is appropriately redundant. An LFSR is such that a bad guess is weeded out with very high likelihood using very simple tests, unlikely to falsely rule out a bad guess. We can even do with a lenient chi-squared test that the plaintext resulting from a guess is random, and when it's not, it will be correct. This has the advantage of working regardless of language.

As long as enough bits can be guessed, these techniques work with LFSRs of hundreds of bits, including with unknown feedback polynomial (just add the taps as additional unknowns).