Let the random variable $X$ follow the distribution:
$$ f(x;\theta) = \theta^2(x+1)(1-\theta)^x$$
where $x$ takes values in $[0, \infty)$ and $\theta$ in $[0, 1]$. The likelihood is defined as:
$$\mathcal{L}(x \mid \theta) = p_\theta(x)$$
and for $N$ i.i.d observations:
$$\mathcal{L}(\theta) = \prod_{i=1}^N p_\theta(x_i)$$
Due to the monoticity of the $\ln$ function, we can instead maximize the log-likelihood:
$$\ln(\mathcal{L}(\theta)) = \ln \left( \prod_{i=1}^N p_\theta(x_i) \right) = \sum_{i=1}^{N} \ln p_\theta(x_i) = \sum_{i=1}^{N} x_i\ln \left[\theta^2 (x+1)(1-\theta)\right]$$
The derivative of the likelihood with respect to $\theta$ is:
$$\frac{d\ln(\mathcal{L}(\theta))}{d\theta} = \sum_{i=1}^{N} x_i \frac{d}{d\theta} \ln\left[\theta^2 (x+1)(1-\theta)\right] = \sum_{i=1}^{N} x_i \frac{d}{d\theta} \left[ \ln(\theta^2) + \ln(x_i + 1) + \ln(1-\theta)\right]$$
Taking the derivatives of the terms inside the brackets we obtain:
$$\sum_{i=1}^{N} x_i \left(\frac{2}{\theta} - \frac{1}{1-\theta}\right)$$
and setting this to zero leads to:
$$\sum_{i=1}^{N} x_i \frac{2}{\theta} = \sum_{i=1}^{N} x_i \frac{1}{1-\theta} \Longleftrightarrow \frac{2}{\theta}\sum_{i=1}^n x_i = \frac{1}{1-\theta} \sum_{i=1}^n x_i \Longleftrightarrow \frac{2}{\theta} = \frac{1}{1-\theta}$$
Solving the last equations leads to $\hat{\theta}=\frac{2}{3}$.
Is it possible for the MLE to be independent of the observations ($x_i$ are not involved in calculation of $\hat{\theta}$)? If yes, is there something in the form of $f(x;\theta)$ that indicates that?