This is essentially the Frobenius coin problem: what is the largest amount that cannot be obtained using only coins of specified denominations?
In your case, at each step you can either add 7 pieces of paper, or 11 pieces, to the initial single piece that we are start with. So, the allowed denominations (in the language of the Frobenius coin problem) are 7 and 11.
It turns out that a largest unobtainable amount exists when the greatest common divisor of the set of allowed denominations is equal to $1$, that is, they have no common factor. Since $\gcd(7,11) = 1$, the answer to your problem is: yes, there is a maximum number of pieces that cannot be reached.
Furthermore, it turns out that there is an exact formula for the maximum amount that cannot be obtained in the case when there are only two allowed denominations, $x$ and $y$, which is $xy - x - y$. (No formula is known when there are more than two allowed denominations.)
So, when $x = 7$ and $y = 11$, we have $xy - x - y = 7 \cdot 11 - 7 - 11 = 59$. Since we start with a single piece of paper to which we add multiples of $x = 7$ and $y = 11$, the maximum number of pieces that we cannot obtain is $59+1=60$.