Consider random variable Y with a Poisson distribution: P ( y | θ ) =...

The posterior mode is just the maximizing value of the posterior, so this is essentially just a calculus problem. You have already correctly derived the posterior kernel:
$\pi (\theta |\mathbb{y})\propto \exp ((n\overline{y}-1)\ln \theta -n\theta ).$
So the log-posterior can be written as:
$F_{\mathbb{y}}(\theta )\equiv \ln \pi (\theta |\mathbb{y})=(n\overline{y}-1)\ln \theta -n\theta +\text{const}.$
We can maximise this via ordinary calculus techniques. Differentiating with respect to $\theta$ gives:
$\frac{d{F}_{\mathbb{y}}}{d\theta }(\theta )=\frac{n\overline{y}-1}{\theta }-n\frac{d^2{F}_{\mathbb{y}}}{d{\theta }^2}(\theta )=-\frac{n\overline{y}-1}{{\theta }^2}.$
You are told that $n\overline{y}>1$ so the second derivative of the objective function is negative. This means that the objective function is strictly concave, so the maximizing value occurs at the unique critical point:
$0=\frac{d{F}_{\mathbb{y}}}{d\theta }(\hat{\theta })=\frac{n\overline{y}-1}{\hat{\theta }}-n⟹\hat{\theta }=\overline{y}-\frac{1}{n}.$
So in this case we have mode $\text{mode }\pi (\theta |\mathbb{y})=\overline{y}-1/n$ (which is strictly positive since $\overline{y}>1/n$).