For proposition $p$ to be true, the equation $x^{2}+mx+1=0$ must have two distinct negative roots, which implies:
$$\begin{cases} \Delta > 0 \ x_{1}+x_{2} 0\ \end{cases}$$

This simplifies to:
$$$$
$\text{m 2 -4}$ > 0 \
-m 2$.

For proposition $q$ to be true, the inequality $3^{x}-m+1\leqslant 0$ must have real number solutions, which implies $3^{x}\leqslant m-1$. This is only possible when $m-1 > 0$, so solving this inequality gives us $m > 1$.

Since $p\lor q$ is true and $p\land q$ is false, $p$ and $q$ must have opposite truth values. We consider two cases:

1. $p$ is true and $q$ is false. This leads to the system of inequalities:
$$\begin{cases} m > 2 \ m\leqslant 1\ \end{cases}$$
which has no valid values for $m$.

2. $p$ is false and $q$ is true. This leads to the system of inequalities:
$$\begin{cases} m\leqslant 2 \ m > 1\ \end{cases}$$
which simplifies to $1 < m\leqslant 2$.

Thus, the range of real number values for $m$ is $\boxed{(1,2]}$.