Additional comments: Your answer seems OK. It may be of interest to know that
$\hat \theta$ is not unbiased. One can get a rough idea of the distribution
of $\hat \theta$ for a particular $\theta$ by simulating many samples of
size $n.$ I don't know of a convenient 'unbiasing' constant multiple.
The Wikipedia article I linked in my Comment above gives more information.
Here is a simulation for $n = 10$ and $\theta = 5.$
th = 5; n = 10
th.mle = -n/replicate(10^6, sum(log(rbeta(n, th, 1))))
mean(th.mle)
## 5.555069 # aprx expectation of th.mle > th = 5.
median(th.mle)
## 5.172145
The histogram below shows the simulated distribution of $\hat \theta.$
The vertical red line is at the mean of that distribution, and the green
curve is its kernel density estimator (KDE). According to the KDE, its mode is near $4.62.$
den.inf = density(th.mle)
den.inf$x[den.inf$y==max(den.inf$y)]
## 4.624876
hist(th.mle, br=50, prob=T, col="skyblue2", main="")
abline(v = mean(th.mle), col="red")
lines(density(th.mle), lwd=2, col="darkgreen")
Addendum on Parametric Bootstrap Confidence Interval for $\theta:$
In order to find a confidence interval (CI) for $\theta$ based on MLE $\hat \theta,$ we would like to know the distribution of $V = \frac{\hat \theta}{\theta}.$ When that distribution is not
readily available, we can use a parametric bootstrap.
If we knew the distribution of $V,$ then we could find numbers $L$ and $U$ such that
$P(L \le V = \hat\theta/\theta \le U) = 0.95$ so that a 95% CI would be of the form
$\left(\frac{\hat \theta}{U},\, \frac{\hat\theta}{L}\right).$ Because we do not know the distribution of $V$ we use a bootstrap procedure to get serviceable approximations $L^*$ and $U^*$ of $L$ and $U.$ respectively.
To begin, suppose we have a random sample of size $n = 50$ from $\mathsf{Beta}(\theta, 1)$
where $\theta$ is unknown and its observed MLE is $\hat \theta = 6.511.$
Entering, the so-called 'bootstrap world'. we take repeated 're-samples` of size $n=50$
from $\mathsf{Beta}(\hat \theta =6.511, 0),$ Then we we find the bootstrap
estimate $\hat \theta^*$ from each re-sample. Temporarily using the observed
MLE $\hat \theta = 6.511$ as a proxy for the unknown $\theta,$ we find a large number $B$ of re-sampled values $V^* = \hat\theta^2/\hat \theta.$ Then we use quantiles .02 and .97 of
these $V^*$'s as $L^*$ and $U^*,$ respectively.
Returning to the 'real world'
the observed MLE $\hat \theta$ returns to its original role as an estimator, and the
95% parametric bootstrap CI is $\left(\frac{\hat\theta}{U^*},\, \frac{\hat\theta}{L^*}\right).$
The R code, in which re-sampled quantities are denoted by .re instead of $*$, is shown below.
For this run with set.seed(213) the 95% CI is $(4.94, 8.69).$ Other runs with unspecified
seeds using $B=10,000$ re-samples of size $n = 50$ will give very similar values. [In a real-life application, we would not know whether this CI covers the 'true' value of $\theta.$ However,
I generated the original 50 observations using parameter value $\theta = 6.5,$ so in this demonstration we
do know that the CI covers the true parameter value $\theta.$ We could have used the
probability-symmetric CI with quantiles .025 and .975, but the one shown is a little shorter.]
set.seed(213)
B = 10000; n = 50; th.mle.obs=6.511
v.re = th.mle.obs/replicate(B, -n/sum(log(rbeta(n,th.mle.obs,1))))
L.re = quantile(v.re, .02); U.re = quantile(v.re, .97)
c(th.mle.obs/U.re, th.mle.obs/L.re)
## 98% 3%
## 4.936096 8.691692
I see no need to use log likelihood here. The function itself is decreasing in $\theta$.
$$
\mathcal{L}(\theta; X_1,\dots, X_n) = f(X_1)\cdot\ldots\cdot f(X_n)
$$
$$= \begin{cases}\frac{2X_1}{(\theta+1)^2}, & 0\leq X_1\leq \theta+1 \cr 0, & \text{else}\end{cases}\;\times\;\ldots\;\times\; \begin{cases}\frac{2X_n}{(\theta+1)^2}, & 0\leq X_n\leq \theta+1 \cr 0, & \text{else}\end{cases}
$$
$$
=\begin{cases}\frac{2^n\prod\limits_{i=1}^n X_i}{(\theta+1)^{2n}}, & 0\leq X_1,\ldots,X_n\leq \theta+1 \cr 0, & \text{else}\end{cases}
$$
(here $X_{(n)}=\max(X_1,\ldots,X_n)$)
$$
=\begin{cases}\frac{2^n\prod\limits_{i=1}^n X_i}{(\theta+1)^{2n}}, & X_{(n)}\leq \theta+1 \cr 0, & \text{else}\end{cases}
$$
$$
=\begin{cases}\frac{2^n\prod\limits_{i=1}^n X_i}{(\theta+1)^{2n}}, & \theta\geq X_{(n)}-1 \cr 0, & \theta< X_{(n)}-1 \end{cases}
$$
Since $\frac{2^n\prod\limits_{i=1}^n X_i}{(\theta+1)^{2n}}$ decrease as $\theta$ increase, the highest value of $\mathcal{L}(\theta; X_1,\dots, X_n)$ is attained at the smallest value of $\theta$ satisfying the inequality $\theta\geq X_{(n)}-1$. So, MLE is $\hat\theta = X_{(n)}-1$.
Best Answer
The likelihood function, as a function of the two variables $\theta$ and $\sigma$, takes on arbitrarily large values when evaluated at $\theta=x_1$ and at $\sigma$ very close to $0$. Because of the division by $\sigma$ in the second term in your formula for the density. Other values of $\theta$ close to any of the $x_i$ and with $\sigma\ge\min_i|x_i-\theta|$ will similarly give rise to large values of the likelihood.