Your conjecture is true. Here is a proof.


Define $\operatorname{Av}_n(w)$ to be the number of binary words of length $n$ which avoid the pattern $w$.

Let $u$ and $v$ be binary words with $|u| = k$ and $|v|=m$ with $k < m$. We will show that $\operatorname{Av}_n(u) < \operatorname{Av}_n(v)$ for all $n$.

Define the special words
  $$M_n = \overbrace{00\cdots00}^n\qquad\text{and}\qquad L_n = \overbrace{00\cdots01}^n.$$

We can show fairly easily using the cluster method of Goulden and Jackson (and other ways as well, though the cluster method works easily for any pattern) that words avoiding $M_n$ and words avoiding $L_n$ have the generating functions
  $$m_n(x) = \sum_{r \geq 0}\operatorname{Av}_r(M_n)x^r = \frac{1-x^n}{1-2x+x^{n+1}}$$
and
  $$\ell_n(x) = \sum_{r\geq 0} \operatorname{Av}_r(L_n)x^r = \frac{1}{1-2x+x^n}.$$

Moreover, of all words $w$ with $|w|=n$, $M_n$ is the most avoided word and $L_n$ is the least avoided word. Formally, for $w$ with $|w|=n$ and all $r$
  $$\operatorname{Av}_r(L_n) \leq \operatorname{Av}_r(w) \leq \operatorname{Av}_r(M_n).$$

This can be seen probabilistically by observing that the number of occurrences of a pattern of length $n$ in *all* words of length $r$ is independent of what the pattern is. Since $M_n$ "packs" the most easily (i.e., has a lot of overlaps) and $L_n$ does not "pack" at all (i.e., cannot overlap itself), it follows that $M_n$ appears as a pattern in less words overall than any other pattern and $L_n$ appears as a pattern in more words overall than any other pattern.

It should also be obvious that $\operatorname{Av}_r(L_n) \leq \operatorname{Av}_r(L_{n+1})$ for all $r$.

We need to prove one more fact: $\operatorname{Av}_r(M_{s-1}) < \operatorname{Av}_r(L_s)$ for all $r \geq s-1$. We can do this by looking at the generating functions given above. It's clear that the inequality is *eventually* true because the generating functions tell us that exponential growth rate of words avoiding $M_{s-1}$ is strictly less than the exponential growth rate of words avoiding $L_s$. However, we can actually prove this inequality for all $r \geq s-1$ by subtracting $m_{s-1}(x)$ from $\ell_s(x)$ and showing that the result has positive coefficients.

\begin{align*}
  \ell_s(x) - m_{s-1}(x) &= \frac{1}{1-2x+x^s} - \frac{1-x^{s-1}}{1-2x+x^s}\\[6pt]
  &= \frac{x^{s-1}}{1-2x+x^s}\\[6pt]
  &= x^{s-1} \cdot \ell_{s}(x).
\end{align*}

Since $\ell_{s}(x)$ has positive coefficients, our difference has positive coefficients for all powers of $x$ at least $s-1$. This proves the inequality.

We now combine all of our results: for $r \geq k$
  $$\operatorname{Av}_r(u) \leq \operatorname{Av}_r(M_k) < \operatorname{Av}_r(L_{k+1}) \leq \operatorname{Av}_r(L_m) \leq \operatorname{Av}_r(v).\;\;\square$$