k,m,n>0 and p is prime

p^(2k+1)*m*(mn+1)^2 + m^2 is a square ... (1)

Consider a^2+b^2 which can be written as a^2 +b^2+2*a*b-2*a*b
                                         = (a+b)^2 - 2*a*b ...(2)

Using (2) on expression (1) we can transform the given expression to

(sqrt(p^(2k+1)*m*(mn+1)^2) + m)^2 - 2*sqrt(p^(2k+1)*m*(mn+1)^2)*m ...(3)

Since expression (3) is given to be a perfect square, we can say the following
(this is where I suspect I am not totally correct):

2*sqrt(p^(2k+1)*m*(mn+1)^2)*m = 0     ...(4)

=> m = 0 or mn+1 = 0
=> since m,n>0 hence, mn+1 > 0 always and m != 0

So according to my proof, there is no m which satisfies these
conditions of the given problem statement. (I don't know how to use LaTeX
here on this editor; any help is greatly appreciated)