Does every odd integer $m$ satisfy $3^x(m)-2^y=1$ for some integer values of $x$ and $y$?

No, this is not true in general.

For $m$ a power $3^n$ of $3$, we can rewrite the equation as $$3^{x + n} - 2^y = 1,$$ but then it follows from Mihăilescu's Theorem that solutions are only possible for $n \leq 2$, giving for $m = 1, 3$, respectively the solutions $(2, 3)$ and $(1, 3)$.

This is not the only obstacle: Reducing the equation modulo $m$ and rearranging leaves $$2^y \equiv -1 \pmod m ,$$ but this congruence only admits a solution if $2$ has even order in the group $(\Bbb Z / m \Bbb Z)^\times$ of units modulo $m$. This means there are no solutions for $m = 7, 15, 21, 23, 31, \ldots$, that is, for the elements of OEIS A014659.

There are yet other examples: For example, for $m = 13$, $2$ has order $12$ and so the above congruence implies $y = 12 z + 6$, and the equation becomes $$13 \cdot 3^x - 2^6 \cdot (2^{12})^z = 1 .$$ Reducing modulo $4$ gives $(-1)^x \equiv 1 \pmod 4$, so $x = 2 a$, and the equation becomes $$13 \cdot 9^a - 2^6 \cdot (2^{12})^z = 1 .$$ Finally, reducing modulo $5$ and rearranging leaves $3 \cdot (-1)^a \equiv 0 \pmod 5$, but this has no solutions.

On the other hand, we observe that for $m = 11$, $x = 1, y = 5$ is a solution. Together with Conrad's observation in the comments, this shows that the only odd values $m$, $1 \leq m \leq 15$, that admit solutions are $m = 1, 3, 11$.

A quick computer search finds that the only other $m < 1\,000$ with solutions $(x, y)$ with $x < 1\,000$ are $19, 43, 57, 171, 683$.

Edit In fact, the answer to a question motivated by this one shows that the $m$ that admit solutions are precisely those of the form $$m = \frac{2^{3^{y - 1} (2 k + 1)} + 1}{3^y} ,$$ and their corresponding solutions are $$(3^{y - 1} (2 k + 1), y) .$$

No, take $m$ to be a power of $3$, your question reduces to: Distance between powers of 2 and 3

Hope it helps:)