Solution:

Yes, there are infinitely many such rows. For example,
$$\binom{203}{68} = 2\binom{203}{67} \text{ and } \binom{203}{85} = 2\binom{203}{83}$$
There are infinitely many rows having two adjacent elements in a 1:2 ratio, for
$$2\binom{n}{k} = \binom{n}{k+1}$$
reduces to $2(k+1) = n - k$, or $n = 3k + 2$. So as long as $n \equiv 2 \pmod{3}$, there will be two adjacent elements in a 1:2 ratio.

Next, we search for "doubles" that are not adjacent. The next easiest case to try is
$$2\binom{n}{k} = \binom{n}{k+2}$$
which reduces to
$$2(k+2)(k+1) = (n-k)(n-k-1).$$
Substitute $u = n - k$ and $v = k + 2$; our equation becomes
$$2(v^2 - v) = u^2 - u$$
Multiplying both sides by 4 and completing the square yields
$$2(4v^2 - 4v + 1) = 4u^2 - 4u + 1 + 1,$$
so substituting $x = 2v - 1$, $y = 2u - 1$ reduces the original equation to
$$2x^2 - y^2 = 1.$$
This is a Pell's equation with infinitely many solutions which can be generated in the standard way, or we can observe that $(1,1)$, $(5,7)$ are solutions and that if $(x, y)$ is a solution, then $(3x + 2y, 4x + 3y)$ is also a solution.

It remains to show that one of these solutions produces an $n$ which is congruent to $2$ modulo $3$. Since $n = \frac{x + y}{2} - 1$, we must have $x + y \equiv 0 \pmod{6}$. The first solution with this property is $(5,7)$, but this doesn't work, because it corresponds to $n = 5$, and the elements of the row are $1, 5, 10, 10, 5, 1$, so the doubles are not distinct. Generating solutions via $(x, y) \rightarrow (3x + 2y, 4x + 3y)$ modulo $6$, we have the repeating pattern
$$(1,1), (-1,1), (-1,-1), (1,-1), (1,1), \ldots$$
which shows that every other solution $(x, y)$ has the property $x + y \equiv 0 \pmod{6}$. The next solution after $(5,7)$ with this property will correspond to an $n$ large enough so that the doubles will be distinct. Specifically, the solution is $(169, 239)$ which corresponds to $n = 203$, $k = 83$.