The OP's specific Diophantine problem has been answered, but it's still worth pointing out a general technique that applies to problems of this kind:

Given an algebraic function $f(x)$, find all $x$ such that both $x$ and $f(x)$ are integers

whenever $f$ is not itself a polynomial but can be expanded in a power series about $x=\infty$ **with rational coefficients**. Such is the case for the function
$$
f(m) = \frac{m (7m^2-22m+7)}{\sqrt{(m^2-2m+9)(9m^2-2m+1)}}
= \frac{7}{3} m - \frac{128}{27} + O(1/m)
$$
of the combinatorics problem, or indeed its denominator $$\sqrt{(m^2-2m+9)(9m^2-2m+1)} = 3m^2−\frac{10}{3}m+\frac{337}{27}+O(1/m)$$ (as I noted in my comment to C.Matthew's answer). Often $f$ is a rational function, which automatically satisfies the condition as long as it is not a polynomial. The technique is simply:

Use the power-series expansion to write $f(x) = P(x) + O(1/x)$ with $P \in {\bf Q}[x]$, find a common denominator $D$ so that $DP \in {\bf Z}[x]$, and observe that once $x$ is large enough that the $O(1/x)$ error drops below $1/D$ in absolute value the only way that $Df(x)$ can be integral is for $x$ to be a root of $f(x) = P(x)$. This reduces the problem to a finite search.

The OP didn't say where exactly his question came from, but such problems arise often in the theory of combinatorial designs and strongly regular graphs. For example, in a Moore graph of girth 5 and degree $d$, every vertex has $d$ neighbors and any two distinct vertices are xeither adjacent xor have a common neighbor, in which case that neighbor is unique. Using the fact that every eigenvalue of the adjacency matrix has integral multiplicity, one shows that either $d=2$ or $d=m^2+m+1$ for some integer $m$ such that $(m^2+m+1)(m^2+m-1) / (2m+1) \in {\bf Z}$. So we write
$$
16 \frac{(m^2+m+1) (m^2+m-1)} {2m+1} = 8m^3+12m^2+2m-1 - \frac{15}{2m+1}
$$
and (since $15/(2m+1)$ cannot vanish) deduce that $2m+1 \leq 15$, so $m \leq 7$. We then find that of the remaining candidates only $m=1,2,7$ work. Hence $d$ is one of 2, 3, 7, or 57. [See for instance Cameron and Van Lint's *Designs, Graphs, Codes and their Links* (LMS 1991, 1996) for this and many more examples. As it happens, each of $d=2,3,7$ occurs for a unique Moore graph, namely the 5-cycle, Petersen graph, and Hoffman-Singleton graph respectively, while the existence of a Moore graph of degree 57 is a famous open problem.]

Further shortcuts may be available when $D$ and/or the implicit constant in $O(1/x)$ are large enough that the concluding search, though finite, is still inconvenient or infeasible to do exhaustively. For example, in the Moore graph problem, $15/(2m+1)$ must be integral, so $2m+1$ must be a factor of $15$, and this immediately yields the solutions $m=1,2,7$, corresponding to factors $3,5,15$. The numerator $15$ could have been predicted by from the value $15/16$ taken by $(m^2+m+1) (m^2+m-1)$ at $m=-1/2$, the root of the denominator $2m+1$. More generally if $f(x)$ is a rational function $A(x)/B(x)$ with $A,B \in {\bf Z}[x]$ relatively prime, we can use the Euclidean algorithm for polynomials to find $M,N \in {\bf Z}[x]$ such that
$MA-NB$ is a nonzero integer, say $R$; then if $f(x)$ is an integer then so is
$$
M(x) f(x) - N(x) = \frac{M(x)A(x) - N(x)B(x)}{B(x)} = \frac{R}{B(x)} ,
$$
and we need only find integer solutions of $B(x) = r$ for each of the factors $r$ of $R$ and test each of these candidates. $R$ is a factor of the resultant of $A$ and $B$. This is actually not far from G.Robinson's analysis: if we square to get the rational function $m^2 (7m^2-22m+7)^2 / ((m^2-2m+9)(9m^2-2m+1))^2$, the resultant is $2^{36} 3^{12}$, explaining the factors of 2 and 3 that arose in that solution. There are still $2 \cdot 37 \cdot 13 = 962$ factors to try (allowing negative as well as positive $r$), but that's much less than the number of candidates that the generic method would require testing.