In many cases there's a relatively simple, mindless test you can apply.
Recall that an Exponential family of distributions has densities of the form
$$f(x,\theta) = \exp(\eta(\theta)T(x) + A(\theta)+B(x)).$$
Suppose there is a region of values of $(x,\theta)$ in which $\eta,$ $T,$ $A,$ and $B$ are differentiable. Applying the logarithm and taking derivatives shows
$$\frac{\partial^2}{\partial x\partial \theta}\log f(x,\theta) = \eta^\prime(\theta)T^\prime(x),\tag{*}$$
effectively "killing off" the $A$ and $B$ terms.
If you can restrict this region to one where $\eta^\prime$ and $T^\prime$ each remain positive or negative, without equaling zero, and each is differentiable, you can repeat this process (after taking absolute values, if necessary, to assure the log can be applied):
$$\frac{\partial^2}{\partial x\partial \theta} \log|\frac{\partial^2}{\partial x\partial \theta}\log f(x,\theta)| = 0.$$
The result is zero because taking the log splits $(*)$ into a sum of a function of $\theta$ and a function of $x;$ just as at the outset, the mixed partial derivative kills both terms.
The Lindley-Poisson distribution is a discrete distribution on the values $x\in\{0,1,2,\ldots\}$ with probabilities
$$f(x,\theta) = \frac{\theta^2(x+\theta+2)}{(\theta+1)^{x+3}}$$
for $\theta\gt 0,$ giving
$$\log f(x,\theta) = 2\log\theta + \log(x+\theta+2) - (x+3)\log(\theta+1).$$
Its mixed partial derivative can be mechanically computed using basic laws of differentiation as
$$\frac{\partial^2}{\partial x\partial \theta}\log f(x,\theta) = 0 - \frac{1}{(x+\theta+2)^2} - \frac{1}{\theta+1}.$$
This is constantly negative. Repeating this operation on its absolute value gives (again purely mechanically)
$$\frac{\partial^2}{\partial x\partial \theta} \log|\frac{\partial^2}{\partial x\partial \theta}\log f(x,\theta)| = \frac{6}{(x+\theta+2)^4}\ne 0,$$
proving this is not an Exponential family.
