To achieve that level of confidence, you would have to satisfy the most stringent critic. They would require you to establish
Your sample truly is a simple random sample.
The respondents answer honestly and correctly.
That if the barest minority, just less than half, of the population actually would answer "yes", you would have less than a $100 - 90\% = 10\%$ chance of observing at least this many yeses in such a random sample.
You address $(1)$ by explaining and documenting your procedures to identify the population and obtain a sample from it.
You address $(2)$ by documenting how the questioning was carried out and including additional questions to assess reliability and internal consistency.
You address $(3)$ by computing the chance of observing $14$ or more yeses in a random sample when at most $50\%$ of the population would answer yes if asked. Let's do that.
Assuming the population is large (any larger than a few hundred would be fine), the distribution of the yes counts would be very close to Binomial with parameters $22$ and $p \lt 50\%$. In the worst case you have to deal with, take $p=50\%$.
Here is a partial chart of the relevant chances. The top row is a threshold count; below it is the chance that the count in a sample would equal or exceed it.
Threshold: 11 12 13 14 15 16 17 18 19 20 21 22
Chance: 0.58 0.42 0.26 0.14 0.07 0.03 0.01 0.00 0.00 0.00 0.00 0.00
Since the chance of 14 or more is $0.14=14\%$ and that's greater than $10\%$, you cannot have $90\%$ confidence that a majority of the population would answer yes.
If the population is much smaller than several hundred, the confidence in these results noticeably increases. In the very best case, where the population is $28$ or smaller, your sample already contains half or more of the yes-responders and your confidence is $100\%$.
R calculation of the table was
x <- 11:22
y <- round(pbinom(x-1, 22, 1/2, lower.tail=FALSE), 2)
names(y) <- x
Alternatively, you could use the Normal approximation to this Binomial distribution and compute instead
y <- round(pnorm(x-1+1/2, 11, sqrt(22*1/2*(1-1/2)), lower.tail=FALSE), 2)
To two decimal places the results are the same. With even less computation you could find the critical threshold approximately as
ceiling(qnorm(0.90, 11, sqrt(22*1/2*(1-1/2))) + 1/2), which returns $15$.