Confused about Two Sample Proportion Tests - am I performing this correctly? I am dealing with an advertising dataset that includes hundreds of advertisements, how many clicks each ad receives, whether the amount of clicks received is considered "high" or "low" (determined by a methodology I don't have access to), and whether each ad includes the logo of the advertiser.
I am currently running a Two Tailed T-Test, but would also like to run a Two Sample Proportion Test to determine whether ads with the logo have a higher proportion of "high" click ads than ads without the logo. However, I am a bit confused on the formula to calculate Z Score.
I would really appreciate if somebody could let me know if the below methodology is accurate.

Also, please let me know if there are better statistical tests to run for a dataset like this. Looking for any and all suggestions! Thank you!
 A: I can't read the computations you posted. This seems to be a test whether two binomial proportions are significantly different. To compute the denominator, two approaches are in common use.
[P-values below are for two-sided tests, of $H_0: p_1=p_2$ against $H_a: p_1 \ne p_2.$]
(a) Pool together observations from both groups, and successes from both groups, use
the ratio to estimate standard error. Rationale: under $H_0: p_1 = p_2.$
This method is shown here.
(b) Find the two $\hat p_i$s separately, use them to get the total
variance of $\hat p_1 - \hat p_2,$ and use that to estimate the standard error. Rationale: Separate standard errors are required for the CI, so
the CI 'matches' the test.
Below is output from Minitab, which gives a choice whether to use (a) or (b). Output is for method (b).
Test and CI for Two Proportions 

Sample    X    N  Sample p
1        18   31  0.580645
2       273  337  0.810089


Difference = p (1) - p (2)
Estimate for difference:  -0.229444
95% CI for difference:  (-0.408126, -0.0507618)
Test for difference = 0 (vs ≠ 0):  
  Z = -2.52  P-Value = 0.012

Fisher’s exact test: P-Value = 0.005

Method (a) gives $z = -3.01$ and P-value 0.003, so I suppose you did computations for (a). [Fisher's exact test is based on a hypergeometric distribution.]
The procedure prop.test in R gives the following results (because of reasonably large sample sizes I did not use a continuity correction):
 prop.test(c(18,273), c(31, 337), cor=F)

        2-sample test for equality of proportions  
        without continuity correction

data:  c(18, 273) out of c(31, 337)
X-squared = 9.0325, df = 1, p-value = 0.002652
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.40812588 -0.05076184
sample estimates:
   prop 1    prop 2 
0.5806452 0.8100890 

The corresponding $|z| = \sqrt{9.0325}=3.0039,$ in close agreement with Minitab's output for (a).
