$(a)\quad n(4n+1)=m(5m+1)\implies (n-m)(4m+4n+1)=m^2, (n-m)(5m+5n+1)=n^2$ $\gcd(m^2,n^2)=(n-m)\gcd(4m+4n+1,5m+5n+1)=n-m\implies \gcd(m,n)^2=n-m$ Thus, $n-m$ is perfect square.

(b) $(n,m)=(38,34)$

Clearly $m>n$. Let $d={\rm gcd}(m,n)$ with $m=dm_1$ and $n=dn_1$. Then, \[ 4dn_1^2 + n_1 = 5dm_1^2 + m_1\implies n_1-m-1 = d(5m_1^2-4n_1^2). \]Now, let $p\mid m_1-n_1$ be a prime. Then, $m_1\equiv n_1\pmod{p}\implies 5m_1^2-4n_1^2\equiv n_1^2\not\equiv 0\pmod{p}$ as $p\nmid n_1$ (otherwise $p\mid (m_1,n_1)$). Consequently, (a) $5m_1^2-4n_1^2=1$ and (b) $d=n_1-m_1$. Hence, $n-m=d^2$ is a square. For finding a solution, it simply suffices to investigate the (negative) Pell's equation $(2n_1)^2-5m_1^2 = -1$ which admits $(n_1,m_1)=(19,17)$ as a solution. Remark. In fact, if $(m_1,n_1)$ is such that $(2n_1)^2-5m_1^2=-1$, then they are trivially coprime. Thus all solutions are parameterized by solutions of the Pell's equation $x^2-5y^2=-1$ (note that if $y$ is even then $x^2\equiv -1\pmod{4}$, not possible; hence $y$ is odd and $x$ is trivially even).