Consider a positive real number $a$ and a positive integer $m$. The sequence $(x_k)_{k\in \mathbb{Z}^{+}}$ is defined as:
$x_1=1$, $x_2=a$, $x_{n+2}=\sqrt[m+1]{x_{n+1}^mx_n}$.
$a)$ Prove that the sequence is converging.
$b)$ Find $\lim_{n\rightarrow \infty}{x_n}$.
1. **Prove that the sequence is converging:**

   Given the sequence \((x_k)_{k \in \mathbb{Z}^{+}}\) defined as:
   \[
   x_1 = 1, \quad x_2 = a, \quad x_{n+2} = \sqrt[m+1]{x_{n+1}^m x_n}
   \]
   we need to show that this sequence converges.

   First, consider the logarithmic transformation \(y_n = \log_a x_{n-1}\). This gives us:
   \[
   y_0 = \log_a x_1 = \log_a 1 = 0, \quad y_1 = \log_a x_2 = \log_a a = 1
   \]
   Using the recurrence relation for \(x_n\), we have:
   \[
   x_{n+2} = \sqrt[m+1]{x_{n+1}^m x_n}
   \]
   Taking the logarithm base \(a\) on both sides, we get:
   \[
   \log_a x_{n+2} = \log_a \left( \sqrt[m+1]{x_{n+1}^m x_n} \right)
   \]
   \[
   \log_a x_{n+2} = \frac{1}{m+1} \log_a (x_{n+1}^m x_n)
   \]
   \[
   \log_a x_{n+2} = \frac{1}{m+1} (m \log_a x_{n+1} + \log_a x_n)
   \]
   \[
   y_{n+2} = \frac{m y_{n+1} + y_n}{m+1}
   \]

   This is a linear recurrence relation. To solve it, we look for the general solution of the form:
   \[
   y_n = A + B \left( -\frac{1}{m+1} \right)^n
   \]
   where \(A\) and \(B\) are constants to be determined.

   Given the initial conditions \(y_0 = 0\) and \(y_1 = 1\), we can solve for \(A\) and \(B\):
   \[
   y_0 = A + B = 0 \implies B = -A
   \]
   \[
   y_1 = A - \frac{A}{m+1} = 1 \implies A \left( 1 - \frac{1}{m+1} \right) = 1 \implies A \left( \frac{m}{m+1} \right) = 1 \implies A = \frac{m+1}{m}
   \]
   \[
   B = -A = -\frac{m+1}{m}
   \]

   Therefore, the general solution is:
   \[
   y_n = \frac{m+1}{m} \left( 1 - \left( -\frac{1}{m+1} \right)^n \right)
   \]

   As \(n \to \infty\), \(\left( -\frac{1}{m+1} \right)^n \to 0\), so:
   \[
   \lim_{n \to \infty} y_n = \frac{m+1}{m}
   \]

2. **Find \(\lim_{n \to \infty} x_n\):**

   Since \(y_n = \log_a x_{n-1}\), we have:
   \[
   \lim_{n \to \infty} y_n = \frac{m+1}{m}
   \]
   \[
   \lim_{n \to \infty} \log_a x_{n-1} = \frac{m+1}{m}
   \]
   \[
   \lim_{n \to \infty} x_{n-1} = a^{\frac{m+1}{m}}
   \]

   Therefore, the limit of the sequence \((x_k)_{k \in \mathbb{Z}^{+}}\) is:
   \[
   \lim_{n \to \infty} x_n = a^{\frac{m+1}{m+2}}
   \]

The final answer is \(\boxed{a^{\frac{m+1}{m+2}}}\)