I suggest you to use an advise to take $\theta =\lambda^k$ (like Lee David Chung Lin said)
but if you assist to use your notation I introduce you one conjugate prior.(you should check identifiability of your model).
$$f(x_1,\cdots,x_n|\lambda)\propto \frac{1}{\lambda^{nk}}
e^{-\frac{1}{\lambda^k}\sum x_{i}^{k}}$$
if we choose a prior $g(\lambda)$ then
$g(\lambda|x_1,\cdots ,x_n)\propto \frac{1}{\lambda^k}
e^{-\frac{1}{\lambda^k}\sum x_{i}^{k}} g(\lambda)$
so conjugate prior should be something like
$g(\lambda)\propto \frac{1}{\lambda^a} e^{-\frac{b}{\lambda^k}}
\hspace{.5cm} \lambda>0$
I found :
$$\int_{0}^{\infty} y^m e^{-b y^k} dy=\frac{\Gamma(\frac{m+1}{k})}{kb^{\frac{m+1}{k}}}$$
(reference: Table of Integrals, Series, and Products, I.S. Gradshteyn and I.M. Ryzhik, page 337)
by choosing $\lambda=1/y$ in this integral
$$\int_{0}^{\infty} \frac{1}{\lambda^{m+2}} e^{-b \frac{1}{\lambda^{k}}} d\lambda=\frac{\Gamma(\frac{m+1}{k})}{cb^{\frac{m+1}{k}}}$$
so you can create a distribution
$$g(\lambda)=NEWG(m,b)= \frac{\frac{1}{\lambda^{m+2}} e^{-b \frac{1}{\lambda^{k}}} }{\frac{\Gamma(\frac{m+1}{k})}{kb^{\frac{m+1}{k}}}} \hspace{.5cm} \lambda>0$$
so posterior will be
$$g(\lambda|x_1,\cdots ,x_n) \propto
\frac{1}{\lambda^{m+nk+2}} e^{-(b+\sum x_{i}^{k}) \frac{1}{\lambda^{k}}} $$
that is $NEWG(m+nk,b+\sum x_{i}^{k})$
another option is reparmetrize $m_2=mk$.
If you use standard notation $\theta$
$$f(x|\theta)=\frac{kx^{k-1}}{\theta} e^{-\frac{x^k}{\theta}}$$
using inverse gamma for prior
$$g(\theta)\propto \frac{1}{\theta^{\alpha+1}} e^{-\frac{\beta}{\theta}}$$
posterior:
$$g(\theta|x_1,\cdots,x_n)\propto f(x_1,\cdots,x_n|\theta) g(\theta)
=\frac{k^n (\prod_{i=1}^{n} x_i)^{k-1}}{\theta^n} e^{-\frac{\sum_{i=1}^{n} x_i^k}{\theta}} \frac{1}{\theta^{\alpha+1}} e^{-\frac{\beta}{\theta}}$$
$$\propto \frac{1}{\theta^{n+\alpha+1}} e^{-\frac{\beta+\sum_{i=1}^{n} x_i^k}{\theta}}$$
that is Inverse gamma with $(n+\alpha,\beta+\sum_{i=1}^{n} x_i^k)$
Best Answer
It's easy to show if you have an integer value for $\alpha$ in the prior.
We have that the posterior is $\textrm{Gamma}(\alpha^* = \alpha+n,\beta^* = \beta+\sum_{i=1}^n x_i)$, which means that the posterior density is given by:
\begin{equation} p[\lambda | \alpha^*, \beta^*] = \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\lambda^{\alpha^*-1}\exp[-\beta^*\lambda] \end{equation}
The likelihood is an exponential density:
\begin{equation} p[x_{new} | \lambda] = \lambda\exp[-\lambda x_{new}] \end{equation}
Multiplying these two together and moving things around gives us:
\begin{equation} \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\lambda^{\alpha^*}\exp[-(\beta^*+x_{neq})\lambda] \end{equation}
Integrating out $\lambda$, we get:
\begin{equation} \begin{split} & \hspace{6mm }\frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\int_{\lambda \in (0,\infty)}\lambda^{\alpha^*}\exp[-(\beta^*+x_{neq})\lambda]\\ & = \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\int_{\lambda \in (0,\infty)}\frac{(\beta^*+x_{neq})}{(\beta^*+x_{neq})}\lambda^{\alpha^*}\exp[-(\beta^*+x_{neq})\lambda]\\& = \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\frac{1}{(\beta^*+x_{neq})}\int_{\lambda \in (0,\infty)}(\beta^*+x_{neq})\lambda^{\alpha^*}\exp[-(\beta^*+x_{neq})\lambda] \end{split} \end{equation}
Which is the $\alpha^*$ moment of an $\textrm{Exp}(\beta^*+x_{neq})$ distribution. This is equivalent to $\frac{\alpha^*!}{(\beta^*+x_{neq})^{\alpha^*}}$; giving us that the previous expression is: \begin{equation} \begin{split} \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\frac{1}{(\beta^*+x_{neq})}\int_{\lambda \in (0,\infty)}(\beta^*+x_{neq})\lambda^{\alpha^*}\exp[-(\beta^*+x_{neq})\lambda] & = \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\frac{\alpha^*!}{(\beta^*+x_{neq})^{\alpha^*+1}} \end{split} \end{equation}
The $\frac{\alpha^*!}{\Gamma(\alpha^*)}$ term reduces to $\alpha^*$, giving us that the expression reduces to:
\begin{equation} \begin{split} \frac{(\beta^*)^{\alpha^*}}{\Gamma(\alpha^*)}\frac{\alpha^*!}{(\beta^*+x_{new})^{\alpha^*+1}} & =\frac{\beta^{\alpha^*}\alpha^*}{(\beta^*+x_{new})^{\alpha^*+1}} \end{split} \end{equation}
Which matches the density given here with $\lambda = \beta^*$ and $\alpha = \alpha^*$.