For the fun of it, I constructed the smallest counter-examples to both claims by hand (well, actually in my head). Here's the results, as well as how I went about finding them.
Claim 1: The seventh member of the series, $1+2+4+8+16+32+64+128+256=511$ is composite, not prime as specified. $511 = 7*73$.
This claim becomes easier to disprove once we recognize that the elements,
$7, 15, 31, 63, 127, 255, 511, ...$
are the Mersenne numbers, $2^n-1$, starting at $n = 3$. A useful fact about Mersenne numbers is that a Mersenne number can only be prime if its exponent, $n$, is prime. This immediately implies that elements $2, 4, 6, 7$ of the given series are composite. Thus we see that element $7$, $511$, must be a counter-example, because it is an composite odd-numbered element. To verify that it is the smallest counter-example, we need only verify that elements $1,3,5$, namely $7, 31, 127$ are prime. They indeed are, so $511$ is the smallest counter-example.
Claim 2: $n = 20, (119,121)$. $119 = 7*17, 121 = 11*11$
To find this, I categorized the possible solutions by the lower prime factor of $6n-1$ and $6n+1$. Clearly, neither value will be $2$ or $3$. First, I tried the pair $(5,7)$.
I calculated that $x \equiv 0 \mod 5$ and $x \equiv -1 \text{ or } 1 \mod 6$ implies $x \equiv 5\text{ or }25 \mod 30$ respectively, so the other member of the pair must equal $7\text{ or }23 \mod 30$. The smallest multiples of 7 satisfying those congruences, other than 7 itself, are $217$ and $203$, respectively.
Since I was not sure that this solution $(203, 205)$ was as small as possible, I repeated the process for the pair of factors $(5, 11)$. Reusing the congruence $7\text{ or }23 \mod 30$ for the multiple of $11$, I found the smallest multiples $187$ and $143$.
At this point, my smallest counter-example was $(143, 145)$. There only remained one case to check: $(119,121)$, because $121$ is the only number with $11$ as its smallest factor under $143$, and I had already checked that the pair $(5,7)$ had no solutions under $(143, 145)$. As it turns out, $(119,121)$ is a solution, so it must be the smallest.