$\def\p{\partial} \require{enclose}$Start with the van der Waals equation $(1)$.
$$\left(P + a\frac{n^2}{V^2}\right)\left(V - bn\right) = nRT\tag1$$
Note that you have a slightly different form of the equation, namely you have used molar volume $V_m$. Divide both sides of $(1)$ with $n$, use $V_m = V/n$, and $n^2/V^2 = 1/V_m^2$ to reach your variant.
Values for critical parameters in terms of $a, b, n, R$
The critical point $(P_c, T_c, V_c)$ is a critical point and an inflection point of the isotherm by definition.$^{[a]}$ Therefore we arrive at two equations;
$$\left(\frac{\p P}{\p V}\right)_{T\ =\ T_c} = 0,\tag2$$
$$\left(\frac{\p^2 P}{\p V^2}\right)_{T\ =\ T_c} = 0.\tag{2'}$$
Equation $(2)$ is a sufficient condition for a critical point (mathematically, a derivative's non-existance works too). Equation $(2')$ gives a necessary condition for an inflection point.
So our first step should be to find the equation $P = P(T, V)$. Just use equation $(1)$.
$$P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\tag{3}.$$
Apply conditions $(2)$ and $(2')$ on $(3)$.
$$\left(\frac{\p P}{\p V}\right)_T = -\frac{nRT}{(V - bn)^2} + 2a\frac{n^2}{V^3} \overset{(2)}{=} 0\tag4$$
$$\left(\frac{\p^2 P}{\p V^2}\right)_T = \frac{2nRT}{(V-bn)^3} - 6a\frac{n^2}{V^4} \overset{(2')}{=} 0\tag{4'}$$
Both $(4)$ and $(4')$ are equal to zero. Thus we may multiply either equation by non-matching real numbers, and still maintain equality: $(4) = (4')$. We will use this to our advantage by multiplying $(4)$ with $2/(V-bn)$ and $(4')$ by $-1$.
$$-\frac{2nRT}{(V - nb)^3} + 4a\frac{n^2}{V^3(V- nb)} \overbrace{=}^{(4)\ =\ (4')} -\frac{2nRT}{(V - nb)^3} + 6a\frac{n^2}{V^4}\tag5$$
$$\frac{2}{V - nb} = \frac{3}{V} \implies 3V - 3nb = 2V \implies \enclose{box}[mathcolor="green"]{V_c = 3nb}\tag{5'}$$
Plug result $(5')$ into equation $(4')$.
$$-\frac{nRT}{(3nb - nb)^2)} + 2a\frac{n^2}{27n^3b^3} = 0 \implies -\frac{RT}{4} + \frac{2a}{27b} = 0 \implies \enclose{box}[mathcolor="red"]{T_c = \frac{8a}{27bR}}\tag6$$
The outcomes $(5')$ and $(6)$ are to be placed into equation $(3)$.
$$P_c = nR\frac{8a}{27bR} : 2nb - a\frac{n^2}{9n^2b^2} \implies \enclose{box}[mathcolor="blue"]{P_c = \frac{a}{27b^2}}\tag7$$
Values for $a, b, R$ in terms of critical parameters
$$b \overset{(5')}{=} \frac{1}{3n}V_c\tag{8a}$$
$$a \overset{(7)}{=} 27b^2P_c \overset{(8a)}{=} \frac{3P_cV_c^2}{n^2}\tag{8b}$$
$$R \overset{(6)}{=} \frac{8a}{27bT_c} \overset{(8a, 8b)}{=} \frac{8P_cV_c}{3nT_c}\tag{8c}$$
Reduced van der Waals equation
Replace the values of $a, b, R$ from equations $(8a)$$-$$(8c)$ into the original equation $(1)$. Immediately after this substitution multiply both sides by $3/(P_cV_c)$. You should have the result
$$\left[\frac{P}{P_c} + 3\left(\frac{V_c}{V}\right)^2\right]\left(3\frac{V}{V_c} - 1\right) = 8\frac{T}{T_c}.\tag9$$
Define $P/P_c = \pi, V/V_c = \varphi, T/T_c = \tau$. Finally, enjoy the reduced van der Waals equation
$$\enclose{box}[mathcolor="orange"]{\left(\pi + \frac{3}{\varphi^2}\right)(3\varphi - 1) = 8\tau}.\tag{10}$$
- What does equation $(10)$ imply by there being only reduced quantities $\pi, \varphi, \tau$?
Hint: theorem of corresponding states.
$[a]$ While I believe this to be technically true, I also think it deserves a question of its own. The fact that a phase critical point physically always has to (?) be a inflection point confuses me a bit, so is best left to another answerer.
