The Bayesian test for your question is based on the integrated (rather than maximised) likelihood. So for Poisson we have:

$$\begin{array}{c|c}
H_{1}:\lambda_{1}=\lambda_{2} & H_{2}:\lambda_{1}\neq\lambda_{2}
\end{array}
$$

Now neither hypothesis says what the parameters are, so the actual values are nuisance parameters to be integrated out with respect to their prior probabilities.

$$P(H_{1}|D,I)=P(H_{1}|I)\frac{P(D|H_{1},I)}{P(D|I)}$$

The model likelihood is given by:
$$P(D|H_{1},I)=\int_{0}^{\infty} P(D,\lambda|H_{1},I)d\lambda=\int_{0}^{\infty} P(\lambda|H_{1},I)P(D|\lambda,H_{1},I)\,d\lambda$$

$$=\int_{0}^{\infty} P(\lambda|H_{1},I)\frac{\lambda^{x_1+x_2}\exp(-2\lambda)}{\Gamma(x_1+1)\Gamma(x_2+1)}\,d\lambda$$

where $P(\lambda|H_{1},I)$ is the prior for lambda. A convenient mathematical choice is the gamma prior, which gives:

$$P(D|H_{1},I)=\int_{0}^{\infty} \frac{\beta^{\alpha}}{\Gamma(\alpha)}\lambda^{\alpha-1}\exp(-\beta \lambda)\frac{\lambda^{x_1+x_2}exp(-2\lambda)}{\Gamma(x_1+1)\Gamma(x_2+1)}\,d\lambda$$
$$=\frac{\beta^{\alpha}\Gamma(x_1+x_2+\alpha)}{(2+\beta)^{x_1+x_2+\alpha}\Gamma(\alpha)\Gamma(x_1+1)\Gamma(x_2+1)}$$

And for the alternative hypothesis we have:

$$P(D|H_{2},I)=\frac{\beta_{1}^{\alpha_{1}}\beta_{2}^{\alpha_{2}}\Gamma(x_1+\alpha_{1})\Gamma(x_2+\alpha_{2})}{(1+\beta_{1})^{x_1+\alpha_{1}}(1+\beta_{2})^{x_2+\alpha_{2}}\Gamma(\alpha_{1})\Gamma(\alpha_{2})\Gamma(x_1+1)\Gamma(x_2+1)}$$

Now if we assume that all hyper-parameters are equal (not an unreasonable assumption, given that you are testing for equality), then we have an integrated likelihood ratio of:

$$\frac{P(D|H_{1},I)}{P(D|H_{2},I)}=
\frac{(1+\beta)^{x_1+x_2+2\alpha}\Gamma(x_1+x_2+\alpha)\Gamma(\alpha)}
{(2+\beta)^{x_1+x_2+\alpha}\beta^{\alpha}\Gamma(x_1+\alpha)\Gamma(x_2+\alpha)}
$$

Which you can see that the prior information is still very important. We can't set $\alpha$ or $\beta$ equal to zero (Jeffrey's prior), or else $H_{1}$ will always be favored, regardless of the data. One way to get values for them is to specify prior estimates for $E[\lambda]$ and $E[\log(\lambda)]$ and solve for the parameters - this cannot be based on $x_1$ or $x_2$ but can be based on any other relevant information. You can also put in a few different (reasonable) values for the parameters and see what difference it makes to the conclusion. The numerical value of this statistic tells you how much the data and your prior information about the rates in each hypothesis support the hypothesis of equal rates. This explains why the likelihood ratio test is not always reliable - because it essentially ignores prior information, which is usually equivalent to specifying Jeffrey's prior. Note that you could also specify upper and lower limits for the rate parameters (this is usually not too hard to do given some common sense thinking about the real world problem). Then you would use a prior of the form:

$$p(\lambda|I)=\frac{I(L<\lambda<U)}{\log\left(\frac{U}{L}\right)\lambda}$$

And you would be left with a similar equation to that above but in terms of incomplete, instead of complete gamma functions.

For the binomial case things are much simpler, because the non-informative prior (uniform) is proper. The procedure is similar to that above, and the integrated likelihood for $H_{1}:p_{1}=p_{2}$ is given by:

$$P(D|H_{1},I)={n_1 \choose x_1}{n_2 \choose x_2}\int_{0}^{1}p^{x_1+x_2}(1-p)^{n_1+n_2-x_1-x_2}\,dp$$
$$={n_1 \choose x_1}{n_2 \choose x_2}B(x_1+x_2+1,n_1+n_2-x_1-x_2+1)$$
And similarly for $H_{2}:p_{1}\neq p_{2}$
$$P(D|H_{2},I)={n_1 \choose x_1}{n_2 \choose x_2}\int_{0}^{1}p_{1}^{x_1}p_{2}^{x_2}(1-p_{1})^{n_1-x_1}(1-p_{2})^{n_{2}-x_{2}}\,dp_{1}\,dp_{2}$$
$$={n_1 \choose x_1}{n_2 \choose x_2}B(x_1+1,n_1-x_1+1)B(x_2+1,n_2-x_2+1)$$

And so taking ratios gives:

$$\frac{P(D|H_{1},I)}{P(D|H_{2},I)}=
\frac{B(x_1+x_2+1,n_1+n_2-x_1-x_2+1)}
{B(x_1+1,n_1-x_1+1)B(x_2+1,n_2-x_2+1)}
$$
$$=\frac{{x_1+x_2 \choose x_1}{n_1+n_2-x_1-x_2 \choose n_1-x_1}(n_1+1)(n_2+1)}{{n_1+n_2 \choose n_1}(n_1+n_2+1)}$$

And the choose functions can be calculated using the hypergeometric($r$,$n$,$R$,$N$) distribution where $N=n_1+n_2$, $R=x_1+x_2$, $n=n_1$, $r=x_1$

And this tells you how much the data support the hypothesis of equal probabilities, given that you don't have much information about which particular value this may be.

## Best Answer

I'll show you how the invariance property of the MLE applies to this case. Consider the joint distribution of the two random variables, which depends on the parameter vector $\mathbf{p} = (p_1,p_2)$. Suppose you find the MLE of this parameter vector via the joint distribution, and let's denote this MLE as:

$$\hat{\mathbf{p}}_\text{MLE} = (\hat{p}_1, \hat{p}_2).$$

Now, let's define the function $\tau$ so that $\tau(a,b) = a-b$. Applying this function to both the underlying parameter vector, and its MLE we get:

$$\begin{align} \tau(\mathbf{p}) = \tau(p_1, p_2) &= p_1 - p_2, \\[6pt] \tau(\hat{\mathbf{p}}_\text{MLE}) = \tau(\hat{p}_1, \hat{p}_2) &= \hat{p}_1 - \hat{p}_2. \\[6pt] \end{align}$$

Now, the invariance property of the MLE says that the MLE of $\tau(\mathbf{p})$ is $\tau(\hat{\mathbf{p}}_\text{MLE})$, which is equivalent to saying that the MLE of $p_1 - p_2$ is $\hat{p}_1 - \hat{p}_2$. (Note that your broader assertion that the joint distribution depends on $\mathbf{p}$ only through $p_1-p_2$ is false.)