There is, there's not!

Consider $4n^{2} + 4n - 63m^{2} - 84m = 26$

RHS is even: to make LHS even, m must be even.

Now, if m is even... LHS is a multiple of 4, whereas RHS is not divisible by 4.

Hence, no solution possible!

Reducing the equation modulo $4$ gives us $m^2\equiv 2\pmod 4$ But we know that $2$ is not a quadratic residue modulo $4$ . Therefore,no such pair exists.

Motivation: Mainly,I tried to complete the square but unfortunately couldn't manage something good.But I noticed that the equation contains many squares and most of the coefficients are divisible by $4$ . This motivated me to try congruence and (unless I am very much mistaken) I came up with this solution.

By completing the square, you will notice that the equation is of the pell form x^2 - 7y^2 = -1 in which it has no integer solutions because by modulo 7, x^2 is congruent to 6 in which there is no integral solutions.

4n^2 + 4n- 63m^2 - 84m = 26 ,4n^2 - 4m^2 + 4n -4m = 59m^2 +80m + 26 ,4(n-m)(n+m+1) = 118^2/(118m-99)(118m-61) +5 ,4(n-m)(n+m+1) -5 = 118^2/(118m-99)(118m-61) -(1) but m is integer so from equation (1) 4/(118m-99)(118m-61) which is not possible so sum of integral values of m & n is 0